{"id":13676,"date":"2018-04-10T10:00:15","date_gmt":"2018-04-10T04:30:15","guid":{"rendered":"http:\/\/mycbseguide.com\/blog\/?p=13676"},"modified":"2025-10-09T17:07:11","modified_gmt":"2025-10-09T11:37:11","slug":"cbse-sample-papers-class-11-chemistry","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-11-chemistry\/","title":{"rendered":"CBSE Sample Papers Class 11 Chemistry 2025"},"content":{"rendered":"<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_76 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\" style=\"cursor:inherit\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"Toggle Table of Content\"><span class=\"ez-toc-js-icon-con\"><span class=\"\"><span class=\"eztoc-hide\" style=\"display:none;\">Toggle<\/span><span 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href=\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-11-chemistry\/#CBSE_Class_11_Chemistry_Model_Papers\" >CBSE Class 11 Chemistry Model Papers<\/a><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"CBSE_Sample_Papers_Class_11_Chemistry_2024-25\"><\/span><strong>CBSE Sample Papers Class 11 Chemistry (2024-25)<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>At <strong>myCBSEguide<\/strong>, we offer <strong>free downloads<\/strong> of <strong>CBSE <a href=\"https:\/\/mycbseguide.com\/course\/cbse\/2877\/\">Class 11<\/a> Chemistry sample papers<\/strong> for the <strong>2025 academic year<\/strong> in <strong>PDF format<\/strong>. These sample papers are designed based on the <strong>latest CBSE syllabus<\/strong> and <strong>marking scheme<\/strong> for the current academic session, ensuring that students are fully prepared for their upcoming exams. To score well in Chemistry, students can download <strong>CBSE Sample Papers Class 11 Chemistry<\/strong> from myCBSEguide, which offers free access to high-quality papers.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"Why_Practice_with_Class_11_Chemistry_Sample_Papers\"><\/span><strong>Why Practice with Class 11 Chemistry Sample Papers?<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Practicing with <strong>CBSE sample papers<\/strong> is a key strategy to improve your understanding and exam readiness. The sample papers provided by <strong><a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide\">myCBSEguide<\/a> App and <a href=\"https:\/\/mycbseguide.com\/\">myCBSEGuide<\/a> Website<\/strong>\u00a0come with detailed <strong>solutions<\/strong>, making it easier for you to understand the correct approach to solving different types of questions. <strong>CBSE Sample Papers Class 11 Chemistry<\/strong> are available with detailed solutions, helping students to understand the correct approach to solving questions and improve their exam techniques. By solving these papers, you can:<\/p>\n<ul>\n<li><strong>Familiarize yourself<\/strong> with the exam format and question types.<\/li>\n<li><strong>Assess your preparation level<\/strong> and identify weak areas.<\/li>\n<li><strong>Boost your confidence<\/strong> by practicing under timed conditions.<\/li>\n<\/ul>\n<h3><span class=\"ez-toc-section\" id=\"Access_Latest_Class_11_Chemistry_Sample_Papers_2024-25\"><\/span><strong>Access Latest Class 11 Chemistry Sample Papers (2024-25)<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Our <strong>new Class 11 Chemistry sample papers (2024-25)<\/strong> strictly adhere to the <strong>CBSE blueprint<\/strong> for the current academic session, ensuring that you practice the most relevant and updated content. It\u2019s crucial for all students to refer to the <strong>latest syllabus<\/strong> and <strong>marking scheme<\/strong> to align their preparation accordingly. For teachers, the <strong><a href=\"https:\/\/play.google.com\/store\/apps\/details?id=com.examin8.exam\">Examin8<\/a> App and <a href=\"https:\/\/examin8.com\/\">Examin8<\/a> Website<\/strong> is a powerful tool that allows you to create custom mock tests and practice papers with your own name and logo. With <strong>CBSE Sample Papers Class 11 Chemistry<\/strong>, students can test their knowledge of important topics like chemical reactions, organic chemistry, and atomic structure.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"Where_to_Get_Class_11_Chemistry_Sample_Papers\"><\/span><strong>Where to Get Class 11 Chemistry Sample Papers?<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>You can access <strong>Class 11 Chemistry sample papers<\/strong> in two convenient ways:<\/p>\n<ol>\n<li><strong>Download from the myCBSEguide Website<\/strong>: Get free PDFs for instant download.<\/li>\n<li><strong>Download from the myCBSEguide App<\/strong>: The app offers a user-friendly platform where you can access sample papers, solutions, and additional resources for all major subjects, including <strong>Chemistry<\/strong>, <strong>Physics<\/strong>, <strong>Maths<\/strong>, and more.<\/li>\n<\/ol>\n<h3><span class=\"ez-toc-section\" id=\"Start_Your_Exam_Preparation_Today\"><\/span><strong>Start Your Exam Preparation Today!<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Download the <strong>myCBSEguide app<\/strong> now or visit the <strong>myCBSEguide website<\/strong> to access the <strong>latest CBSE sample papers<\/strong> for <strong>Class 11 Chemistry<\/strong> and other subjects. Make the most of your study time with these high-quality resources designed to help you achieve top marks in your exams.<\/p>\n<p style=\"text-align: center;\"><strong>Class 11 &#8211; Chemistry<br \/>\nSample Paper &#8211; 01 (2024-25)<\/strong><\/p>\n<hr \/>\n<p><b>Maximum Marks: 70<br \/>\nTime Allowed: : 3 hours<\/b><\/p>\n<hr \/>\n<p><b>General Instructions:<\/b><\/p>\n<ol style=\"list-style-type: lower-alpha;\" start=\"1\">\n<li>There are 33 questions in this question paper with internal choice.<\/li>\n<li>SECTION A consists of 16 multiple-choice questions carrying 1 mark each.<\/li>\n<li>SECTION B consists of 5 very short answer questions carrying 2 marks each.<\/li>\n<li>SECTION C consists of 7 short answer questions carrying 3 marks each.<\/li>\n<li>SECTION D consists of 2 case-based questions carrying 4 marks each.<\/li>\n<li>SECTION E consists of 3 long answer questions carrying 5 marks each.<\/li>\n<li>All questions are compulsory.<\/li>\n<li>The use of log tables and calculators is not allowed<\/li>\n<\/ol>\n<hr \/>\n<ol style=\"padding-left: 20px; list-style: decimal;\">\n<li style=\"text-align: center; clear: both; display: block;\"><b>Section A<\/b><\/li>\n<li>Consider the chemical reaction gven as,<br \/>\n&#8211;CO (g) + &#8212;\u00a0<span class=\"math-tex\">{tex}H_2{\/tex}<\/span>(g) \u2192 &#8212;\u00a0<span class=\"math-tex\">{tex}C_8{\/tex}<\/span>H<sub>18<\/sub>(l) + &#8212;\u00a0<span class=\"math-tex\">{tex}H_2O{\/tex}<\/span>.<br \/>\nThis equation can be balanced by inserting the following in blank spaces<\/p>\n<div style=\"margin-left: 20px;\">\n<p>a)8,2,1,17<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)8,17,1,8<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)8,8,1,17<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)8,2,8,17<\/p>\n<\/div>\n<\/li>\n<li>In an alpha scattering experiment, few alpha particles rebounded because\n<div style=\"margin-left: 20px;\">\n<p>a)The mass of the atom is concentrated in the centre<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)Positive charge of the atoms very little space<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)All the positive charge and mass of the atom is concentrated in small volume<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)Most of the space in the atom is occupied<\/p>\n<\/div>\n<\/li>\n<li>Consider the reactions given below. On the basis of these reactions find out which of the algebraic relations given in options (i) to (iv) is correct?\n<ol style=\"list-style-type: upper-alpha;\" start=\"1\">\n<li>C (g) + 4 H (g) <span class=\"math-tex\">{tex} \\to {\\text{ C}}{{\\text{H}}_{\\text{4}}}\\left( {\\text{g}} \\right);{\\text{ }}{\\Delta _{\\text{r}}}{\\text{H }} = {\\text{ x kJ mo}}{{\\text{l}}^{-{\\text{1}}}}{\/tex}<\/span><\/li>\n<li>C (graphite,s) + <span class=\"math-tex\">{tex}{\\text{2}}{{\\text{H}}_{\\text{2}}}\\left( {\\text{g}} \\right){\\text{ }} \\to {\\text{ C}}{{\\text{H}}_{\\text{4}}}\\left( {\\text{g}} \\right);{\\text{ }}{\\Delta _{\\text{r}}}{\\text{H }} = {\\text{ y kJ mo}}{{\\text{l}}^{-{\\text{1}}}}{\/tex}<\/span><\/li>\n<\/ol>\n<div style=\"margin-left: 20px;\">\n<p>a)x = y<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)x = 2y<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)x &lt; y<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)x &gt; y<\/p>\n<\/div>\n<\/li>\n<li>Give the number of electrons in the species, <span class=\"math-tex\">{tex}O_2{\/tex}<\/span> and\u00a0<span class=\"math-tex\">{tex}O_2^+{\/tex}<\/span>.\n<div style=\"margin-left: 20px;\">\n<p>a)16 and 8<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)16 and 14<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)16 and 15<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)32 and 16<\/p>\n<\/div>\n<\/li>\n<li>Which of the following always has a negative value?\n<div style=\"margin-left: 20px;\">\n<p>a)Heat of reaction<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)Heat of solution<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)Heat of formation<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)Heat of combustion<\/p>\n<\/div>\n<\/li>\n<li>What will be the Electronic configuration of the element having atomic number 24?\n<div style=\"margin-left: 20px;\">\n<p>a)1s<sup>2<\/sup> 2s<sup>2<\/sup>\u00a02p<sup><span style=\"font-size: 10.8333px;\">6<\/span><\/sup>\u00a03s<sup>2<\/sup>\u00a03p<sup>6<\/sup>\u00a03d<sup>4<\/sup> 4s<sup>1\u00a0<\/sup>4p<sup>1<\/sup><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)1s<sup>2<\/sup> 2s<sup>2<\/sup>\u00a02p<sup>3<\/sup> 3s<sup>2<\/sup>\u00a03p<sup>6<\/sup>\u00a03d<sup>5<\/sup> 4s<sup>1\u00a0<\/sup>4p<sup>3<\/sup><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)1s<sup>2<\/sup> 2s<sup>2<\/sup>\u00a02p<sup>6<\/sup> 3s<sup>2<\/sup>\u00a03p<sup>6<\/sup>\u00a03d<sup>5<\/sup> 4s<sup>1<\/sup><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)1s<sup>2<\/sup> 2s<sup>2<\/sup>\u00a02p<sup><span style=\"font-size: 10.8333px;\">6<\/span><\/sup>\u00a03s<sup>2<\/sup>\u00a03p<sup>3<\/sup>\u00a03d<sup>5<\/sup> 4s<sup>1\u00a0<\/sup>4p<sup>3<\/sup><\/p>\n<\/div>\n<\/li>\n<li>On the reaction<br \/>\n2Ag + 2H<sub>2<\/sub>SO<sub>4<\/sub>\u00a0<span class=\"math-tex\">{tex}\\longrightarrow{\/tex}<\/span>\u00a0Ag<sub>2<\/sub>SO<sub>4<\/sub>\u00a0+ 2H<sub>2<\/sub>O + SO<sub>2<\/sub>\u00a0sulphuric acid acts as an?<\/p>\n<div style=\"margin-left: 20px;\">\n<p>a)a reducing agent<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)an oxidizing agent<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)a catalyst<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)an acid as well as an oxidant<\/p>\n<\/div>\n<\/li>\n<li>The reaction<br \/>\n<span class=\"math-tex\">{tex}C{H_3}CH = C{H_2} + HBr\\xrightarrow{Peroxide} C{H_3}C{H_2}C{H_2}Br{\/tex}<\/span>\u00a0 can be explained by<\/p>\n<div style=\"margin-left: 20px;\">\n<p>a)carbanion formation<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)free-radical mechanism<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)carbocation formation<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)electrophilic substitution<\/p>\n<\/div>\n<\/li>\n<li>Which of the following organic materials damage DNA of our body?\n<div style=\"margin-left: 20px;\">\n<p>a)Tabacco<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)Coal<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)All of these<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)Petroleum<\/p>\n<\/div>\n<\/li>\n<li>In terms of period and group, where would you locate the element with\u00a0Z = 114\u00a0in the modern periodic table?\n<div style=\"margin-left: 20px;\">\n<p>a)7th period and 14th group<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)6th period 14th group<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)6th period and 13th group<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)7th period and 13th group<\/p>\n<\/div>\n<\/li>\n<li>For the reaction at 298 K, 2A + B\u00a0<span class=\"math-tex\">{tex}\\to{\/tex}<\/span> C, <span class=\"math-tex\">{tex}\\Delta{\/tex}<\/span>H = 400 kJ mol<sup>-1\u00a0<\/sup>and <span class=\"math-tex\">{tex}\\Delta {\\text{S}}{\/tex}<\/span> = 0.2 kJ K<sup>-1<\/sup>mol<sup>-1<\/sup>. At what temperature will the reaction become spontaneous considering <span class=\"math-tex\">{tex}\\Delta {\\text{H}}{\/tex}<\/span> and <span class=\"math-tex\">{tex}\\Delta {\\text{S}}{\/tex}<\/span> to be constant over the temperature range.\n<div style=\"margin-left: 20px;\">\n<p>a)3500 K<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)2000 K<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)1500 K<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)2500 K<\/p>\n<\/div>\n<\/li>\n<li>Soda lime is a mixture of\n<div style=\"margin-left: 20px;\">\n<p>a)NaOH + Ca(OH)<sub>2<\/sub><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)NaOH + MgO<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)NaOH + Mg(OH)<sub>2<\/sub><\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)NAOH + CaO<\/p>\n<\/div>\n<\/li>\n<li><strong>Assertion (A):<\/strong> Allyl and benzyl carbonium ions are stable than propyl corbonium ions.<br \/>\n<strong>Reason (R):<\/strong> Electron releasing groups stabilize cabonium ions.<\/p>\n<div style=\"margin-left: 20px;\">\n<p>a)Both A and R are true and R is the correct explanation of A.<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)Both A and R are true but R is not the correct explanation of A.<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)A is true but R is false.<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)A is false but R is true.<\/p>\n<\/div>\n<\/li>\n<li>\n<p class=\"MsoNoSpacing\"><strong>Assertion (A): <\/strong>Trans-2-butene on reaction with Br<sub><span style=\"font-size: 9.5pt;\">2<\/span><\/sub> gives a meso-2,3-dibromobutane.<br \/>\n<strong>Reason (R):<\/strong> The reaction involves the\u00a0<span style=\"font-size: 9.5pt;\">syn<\/span>-addition of bromine.<\/p>\n<div style=\"margin-left: 20px;\">\n<p>a)Both A and R are true and R is the correct explanation of A.<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)Both A and R are true but R is not the correct explanation of A.<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)A is true but R is false.<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)A is false but R is true.<\/p>\n<\/div>\n<\/li>\n<li><strong>Assertion (A):<\/strong>\u00a0Threshold frequency is a characteristic for a metal.<br \/>\n<strong>Reason (R):<\/strong>\u00a0Threshold frequency is the maximum frequency required for the ejection of electrons from the metal surface.<\/p>\n<div style=\"margin-left: 20px;\">\n<p>a)Both A and R are true and R is the correct explanation of A.<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)Both A and R are true but R is not the correct explanation of A.<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)A is true but R is false.<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)A is false but R is true.<\/p>\n<\/div>\n<\/li>\n<li><strong>Assertion (A):<\/strong> Under similar conditions of temperature and pressure, gases combine in a simple ratio of their volume but, not always.<br \/>\n<strong>Reason (R):<\/strong> Gases deviate from ideal behaviour.<\/p>\n<div style=\"margin-left: 20px;\">\n<p>a)Both A and R are true and R is the correct explanation of A.<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>b)Both A and R are true but R is not the correct explanation of A.<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>c)A is true but R is false.<\/p>\n<\/div>\n<div style=\"margin-left: 20px;\">\n<p>d)A is false but R is true.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"Download_the_myCBSEguide_App_for_Comprehensive_Exam_Preparation\"><\/span><strong>Download the myCBSEguide App for Comprehensive Exam Preparation<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>To excel in your <strong>CBSE<\/strong> exams and other competitive exams like <strong>JEE (Main)<\/strong>, <strong>NEET-UG<\/strong>, and <strong>NDA<\/strong>, \u00a0the <strong><a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide\">myCBSEguide<\/a> App and <a href=\"https:\/\/mycbseguide.com\/\">myCBSEGuide<\/a> Website<\/strong>\u00a0is the ultimate study companion. The app provides <strong>complete study material<\/strong> including:<\/p>\n<ul>\n<li><strong>Chapter-wise practice questions<\/strong> and <strong>sample papers<\/strong>.<\/li>\n<li><strong>NCERT solutions<\/strong> for all subjects.<\/li>\n<li><strong>Revision notes<\/strong> for quick and effective review.<\/li>\n<li><strong>Important question papers<\/strong> and <strong>CBSE guess papers<\/strong> to focus your studies.<\/li>\n<\/ul>\n<p>With all these resources available in one app, students can confidently prepare for <strong>CBSE<\/strong> board exams, entrance tests, and other competitive exams, ensuring a structured and effective learning experience.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"Create_Custom_Mock_Tests_with_Examin8com_App_for_Teachers\"><\/span><strong>Create Custom Mock Tests with Examin8.com App for Teachers<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>For <strong>teachers<\/strong>, the\u00a0<strong><a href=\"https:\/\/play.google.com\/store\/apps\/details?id=com.examin8.exam\">Examin8<\/a> App and <a href=\"https:\/\/examin8.com\/\">Examin8<\/a> Website<\/strong>\u00a0is a powerful tool that allows you to create <strong>custom mock tests<\/strong> and practice papers tailored to your students&#8217; needs. With <strong>Examin8<\/strong>, you can:<\/p>\n<ul>\n<li>Design tests with your <strong>own name<\/strong> and <strong>logo<\/strong> for a personalized touch.<\/li>\n<li>Choose from a variety of question formats to create diverse assessments.<\/li>\n<li>Easily distribute these tests to your students to monitor progress.<\/li>\n<\/ul>\n<p>This is an excellent solution for teachers aiming to provide targeted and effective practice for their students, ensuring improved learning outcomes.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"Why_Choose_myCBSEguide_App_and_Examin8_App\"><\/span><strong>Why Choose myCBSEguide App and Examin8 App?<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<ul>\n<li><strong>Comprehensive Study Material<\/strong>: Access NCERT solutions, sample papers, and revision notes for all major subjects.<\/li>\n<li><strong>JEE, NEET, and NDA Prep<\/strong>: In addition to CBSE, get resources for <strong>JEE (Main)<\/strong>, <strong>NEET-UG<\/strong>, and <strong>NDA<\/strong> exams.<\/li>\n<li><strong>Customizable Assessments for Teachers<\/strong>: Teachers can create personalized practice papers with their branding using <strong>Examin8<\/strong>.<\/li>\n<li><strong>Accessible Anytime, Anywhere<\/strong>: Both apps are available for <strong>iOS<\/strong> and <strong>Android<\/strong>, providing easy access on-the-go.<\/li>\n<\/ul>\n<h3><span class=\"ez-toc-section\" id=\"Start_Your_Exam_Preparation_Today-2\"><\/span><strong>Start Your Exam Preparation Today!<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Download the <strong>myCBSEguide.com app<\/strong> now and access high-quality study materials for <strong>CBSE<\/strong>, <strong>JEE<\/strong>, <strong>NEET<\/strong>, and <strong>NDA<\/strong> exams. Teachers, use the <strong>Examin8.com app<\/strong> to craft custom exams for your students and track their progress effectively.<\/p>\n<p style=\"text-align: center;\"><b>Section B<\/b><\/p>\n<\/div>\n<\/li>\n<li>Calculate the solubility of A<sub>2<\/sub>X<sub>3<\/sub> in pure water, assuming that neither kind of ion reacts with water. The solubility product of A<sub>2<\/sub>X<sub>3<\/sub>, K<sub>sp<\/sub> = 1.1 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span> 10<sup>\u201323<\/sup>.<\/li>\n<li>Helium has an electronic configuration of 1s<sup>2 <\/sup>but it\u00a0is placed in p-block in group 18. Explain.<\/li>\n<li>How many moles of methane are required to produce 22g CO<sub>2<\/sub> (g) after combustion?<\/li>\n<li>Explain why\u00a0the system\u00a0are\u00a0not aromatic.<br \/>\n<img decoding=\"async\" style=\"width: 63px; height: 52px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/dCPY2Hr.png\" alt=\"12\" data-imgur-src=\"dCPY2Hr.png\" \/><\/p>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>What happens when benzene is oxidized at 770K in presence of V<sub>2<\/sub>O<sub>5<\/sub>? Give chemical equation.<\/li>\n<li>In an atom, an electron is moving with a speed of 600 ms<sup>-1<\/sup> with an accuracy of 0.005 %. Find the certainty with which the position of the electron can be located. (h = 6.6 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a010<sup>-34<\/sup> kg m<sup>2<\/sup>s<sup>-1<\/sup>, mass of electron m<sub>e<\/sub> = 9.1\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a010<sup>-31<\/sup> kg).<\/li>\n<li style=\"text-align: center; clear: both; display: block;\"><b>Section C<\/b><\/li>\n<li>Draw the resonating structure of\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>ozone molecule<\/li>\n<li>nitrate ion<\/li>\n<\/ol>\n<\/li>\n<li>Answer:\n<ol class=\"inner-group-question-ol\">\n<li>Consider the same expansion, but this time against a constant external pressure of 1 atm.<\/li>\n<li>At what temperature entropy of a substance is zero?<\/li>\n<li>Predict the change in internal energy for an isolated system at constant volume.<\/li>\n<\/ol>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>Why standard entropy of an elementary substance is not zero whereas standard enthalpy of formation is taken as zero?<\/li>\n<li>Under what conditions will the reaction occur, if\n<ol style=\"list-style-type: lower-alpha;\" start=\"1\">\n<li>both\u00a0<span class=\"math-tex\">{tex}\\Delta{\/tex}<\/span>H\u00a0and <span class=\"math-tex\">{tex}\\Delta{\/tex}<\/span>S are positive<\/li>\n<li>both\u00a0<span class=\"math-tex\">{tex}\\Delta{\/tex}<\/span>H and\u00a0<span class=\"math-tex\">{tex}\\Delta{\/tex}<\/span>S\u00a0are negative<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<\/li>\n<li>Write the anode reaction, the cathode reaction and the net cell reaction in the following cells. Which electrode would be positive terminal in each cell?\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>Zn(s) | Zn<sup>2+ <\/sup>|| Br<sub>2<\/sub> | Br<sup>&#8211;<\/sup>\u00a0| Pt(s)<\/li>\n<li>Cr(s) | Cr<sup>3+<\/sup> || I<sub>2<\/sub> | I<sup>&#8211;<\/sup>\u00a0| Pt(s)<\/li>\n<li>Pt(s) | H<sub>2<\/sub>(g) | H<sup>+<\/sup> (aq) || Cu<sup>2+<\/sup> | Cu(s)<\/li>\n<\/ol>\n<\/li>\n<li>Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represeted as v = 3.29 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a010<sup>15<\/sup> (Hz) [1\/3<sup>2<\/sup> &#8211; 1\/n<sup>2<\/sup>] Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum.<\/li>\n<li>Among the elements B, AI, C and Si\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>Which has the highest first ionization enthalpy?<\/li>\n<li>Which has the most negative electron gain enthalpy?<\/li>\n<li>Which has the largest atomic radius?<\/li>\n<li>Which has the most metallic character?<\/li>\n<\/ol>\n<\/li>\n<li>Calculate the molecular mass of the following.\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>H<sub>2<\/sub>O<\/li>\n<li>CO<sub>2<\/sub><\/li>\n<li>CH<sub>4<\/sub><\/li>\n<\/ol>\n<\/li>\n<li style=\"text-align: center; clear: both; display: block;\"><b>Section D<\/b><\/li>\n<li><strong>Read the following text carefully and answer the questions that follow:<\/strong><br \/>\nOnce an organic compound is extracted from a natural source or synthesised in the laboratory, it is essential to purify it. Various methods used for the purification of organic compounds are based on the nature of the compound and the impurity present in it. Finally, the purity of a compound is ascertained by determining its melting or boiling point. This is one of the most commonly used techniques for the purification of solid organic compounds. In crystallisation Impurities, which impart colour to the solution are removed by adsorbing over activated charcoal. In distillation Liquids having different boiling points vaporise at different temperatures. The vapours are cooled and the liquids so formed are collected separately. Steam Distillation is applied to separate substances which are steam volatile and are immiscible with water. Distillation under reduced pressure: This method is used to purify liquids having very high boiling points.<\/p>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>Which method can be used to separate two compounds with different solubilities in a solvent?<\/li>\n<li>Distillation method is used to separate which type of substance?<\/li>\n<li>Which technique is used to separate aniline from aniline water mixture?<br \/>\n<strong>OR<\/strong><br \/>\nWhy chloroform and aniline are easily separated by the technique of distillation?<\/li>\n<\/ol>\n<\/li>\n<li><strong>Read the following text carefully and answer the questions that follow:<\/strong><br \/>\nThe molecular orbital theory is based on the principle of a linear combination of atomic orbitals. According to this approach when atomic orbitals of the atoms come closer, they undergo constructive interference as well as destructive interference giving molecular orbitals, i.e., two atomic orbitals overlap to form two molecular orbitals, one of which lies at a lower energy level (bonding molecular orbital). Each molecular orbital can hold one or two electrons in accordance with Pauli&#8217;s exclusion principle and Hund&#8217;s rule of maximum multiplicity. For molecules up to N<sub>2<\/sub>, the order of filling of orbitals is:<br \/>\n<span class=\"math-tex\">{tex}\\sigma(1 s)_{\\sigma}^{*}(1 s), \\sigma(2 s)_{ \\sigma}^{*}(2 s), \\pi\\left(2 p_{x}\\right){\/tex}<\/span>\u00a0<span class=\"math-tex\">{tex}=\\pi\\left(2 p_{y}\\right), \\sigma\\left(2 p_{z}\\right),_{\\pi}^{*}\\left(2 p_{x}\\right){\/tex}<\/span><span class=\"math-tex\">{tex}=\\stackrel{*}{\\pi}\\left(2 p_{y}\\right), \\stackrel{*}{\\sigma}\\left(2 p_{z}\\right){\/tex}<\/span><br \/>\nand for molecules after N<sub>2<\/sub>, the order of filling is:<br \/>\n<span class=\"math-tex\">{tex}\\sigma(1 s)_{\\sigma}^{*}(1 s), \\sigma(2 s)_{ \\sigma}^{*}(2 s), \\sigma\\left(2 p_{z}\\right), \\pi\\left(2 p_{x}\\right){\/tex}<\/span><span class=\"math-tex\">{tex}=\\pi\\left(2 p_{y}\\right),_{\\pi}^{*}\\left(2 p_{x}\\right){\/tex}<\/span><span class=\"math-tex\">{tex}=\\stackrel{*}{\\pi}\\left(2 p_{y}\\right), \\stackrel{*}{\\sigma}\\left(2 p_{z}\\right){\/tex}<\/span><br \/>\nBond order\u00a0=\u00a0<span class=\"math-tex\">{tex}\\frac{1}{2}{\/tex}<\/span>[bonding electrons &#8211; antibonding electrons]\nBond order gives the following information:<\/p>\n<ol style=\"list-style-type: upper-roman;\" start=\"1\">\n<li>If bond order is greater than zero, the molecule\/ion exists otherwise not.<\/li>\n<li>Higher the bond order, higher is the bond dissociation energy.<\/li>\n<li>Higher the bond order, greater is the bond stability.<\/li>\n<li>Higher the bond order, shorter is the bond length.<\/li>\n<\/ol>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>Arrange the following negative stabilities of CN, CN<sup>+<\/sup> and CN<sup>&#8211;<\/sup> in increasing order of bond. (1)<\/li>\n<li>The molecular orbital theory is preferred over valence bond theory. Why? (1)<\/li>\n<li>Ethyne is acidic in nature in comparison to ethene and ethane. Why is it so? (2)<br \/>\n<strong>OR<\/strong><br \/>\nBonding molecular orbital is lowered by a greater amount of energy than the amount by which antibonding molecular orbital is raised. Is this statement correct? (2)<\/li>\n<\/ol>\n<\/li>\n<li style=\"text-align: center; clear: both; display: block;\"><b>Section E<\/b><\/li>\n<li>Attempt any five of the following:\n<ol class=\"inner-group-question-ol\">\n<li>Classify the hydrocarbons according to the carbon-carbon bond.<\/li>\n<li>Explain why alkynes are less reactive than alkenes towards addition of Br<sub>2<\/sub>.<\/li>\n<li>What is electrophile in sulphonation?<\/li>\n<li>Why are Alkenes called olefins?<\/li>\n<li>n-propylmagnesium bromide on hydrolysis gives propane. Is there any other Grignard reagent which also gives propane? If so, give its name, structure and equation for the reaction.<\/li>\n<li>Write IUPAC name:\u00a0<span class=\"math-tex\">{tex}\\mathrm{CH}_{3} \\mathrm{CH}-\\mathrm{C}\\left(\\mathrm{CH}_{3}\\right)_{2}{\/tex}<\/span><\/li>\n<li>Which of the two trans-but-2-ene or trans-pent-2-ene is non-polar?<\/li>\n<\/ol>\n<\/li>\n<li>The value of K<sub>p<\/sub> for the reaction,<br \/>\nCO<sub>2<\/sub>(g) + C(s)\u00a0<span class=\"math-tex\">{tex} \\rightleftharpoons {\/tex}<\/span>\u00a02CO (g)<br \/>\nis 3.0 at 1000 K. If initially, <span class=\"math-tex\">{tex}P_{CO_{2}}{\/tex}<\/span> =\u00a00.48 bar and P<sub>CO<\/sub> = 0 bar and pure graphite is present, calculate the equilibrium partial pressures of CO and CO<sub>2<\/sub>.<\/p>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>One of the reactions that take place in producing steel from iron ore is the reduction of iron (II) oxide by carbon monoxide to give iron metal and CO<sub>2<\/sub>.<br \/>\nFeO(s) + CO(g)\u00a0<span class=\"math-tex\">{tex}\\rightleftharpoons{\/tex}<\/span>\u00a0Fe(s) + CO<sub>2<\/sub>(g); K<sub>p <\/sub>= 0.265 atm at 1050 K<br \/>\nWhat is the equilibrium partial pressures of CO and CO<sub>2<\/sub> at 1050 K if the initial pressures are: P<sub>co\u00a0<\/sub>= 1.4 atm and <span class=\"math-tex\">{tex}P_{CO_2}{\/tex}<\/span> = 0.80 atm?<\/li>\n<li>Answer:\n<ol class=\"inner-group-question-ol\">\n<li class=\"question-list\" style=\"clear: both;\">\n<div style=\"margin-left: 30px;\">\n<ol style=\"list-style: lower-roman !important;\" type=\"i\">\n<li>What is the basic principle of chromatography?<\/li>\n<li>What is the general molecular formula of saturated monohydric alcohols?<\/li>\n<\/ol>\n<\/div>\n<\/li>\n<li class=\"question-list nostyle\" style=\"clear: both;\">\n<div class=\"col-12\" style=\"text-align: center;\"><b>OR<\/b><\/div>\n<div style=\"margin-left: 30px;\">\n<ol style=\"list-style: lower-roman !important;\" type=\"i\">\n<li>What is the basic principle involved in the estimation of nitrogen by Dumas method.<\/li>\n<li>Explain hyperconjugation effect. How does hyperconjugation effect explain the stability of alkenes?<br \/>\n<h3><span class=\"ez-toc-section\" id=\"Download_myCBSEguide_App_for_Complete_Exam_Preparation\"><\/span><strong>Download myCBSEguide App for Complete Exam Preparation<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Looking for a comprehensive solution to prepare for <strong>CBSE<\/strong>, <strong>JEE (Main)<\/strong>, <strong>NEET-UG<\/strong>, or <strong>NDA<\/strong> exams? The <strong><a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide\">myCBSEguide<\/a> App and <a href=\"https:\/\/mycbseguide.com\/\">myCBSEGuide<\/a> Website<\/strong>\u00a0is your one-stop platform for all study materials. With this app, you can access:<\/p>\n<ul>\n<li><strong>Complete study material<\/strong> for all CBSE subjects, including <strong>NCERT solutions<\/strong>.<\/li>\n<li><strong>Chapter-wise practice tests<\/strong> and <strong>sample papers<\/strong> designed to boost your exam readiness.<\/li>\n<li><strong>Revision notes<\/strong> for quick reviews and targeted exam preparation.<\/li>\n<li><strong>CBSE guess papers<\/strong> to help you focus on important topics.<\/li>\n<\/ul>\n<p>Whether you\u2019re preparing for school exams or competitive entrance exams, the <strong>myCBSEguide app<\/strong> equips you with everything you need to succeed. Downloading <strong>CBSE Sample Papers Class 11 Chemistry<\/strong> ensures that students have access to a variety of question types, including multiple-choice questions, long answer questions, and short answer questions.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"Examin8com_App_Create_Custom_Mock_Tests_for_Teachers\"><\/span><strong>Examin8.com App: Create Custom Mock Tests for Teachers<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>For <strong>teachers<\/strong>, the <strong><a href=\"https:\/\/play.google.com\/store\/apps\/details?id=com.examin8.exam\">Examin8<\/a> App and <a href=\"https:\/\/examin8.com\/\">Examin8<\/a> Website<\/strong>\u00a0is a powerful tool that allows you to create <strong>custom mock tests<\/strong> and practice papers tailored to your teaching needs. Here\u2019s how Examin8 helps:<\/p>\n<ul>\n<li>Design tests with your <strong>own name<\/strong> and <strong>logo<\/strong>, adding a personalized touch.<\/li>\n<li>Choose from a variety of question formats to match your curriculum.<\/li>\n<li>Easily create and distribute tests to your students for seamless assessments.<\/li>\n<\/ul>\n<p>Examin8 simplifies the process of creating customized assessments, helping teachers provide targeted practice for students.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"Start_Preparing_Today_with_myCBSEguide_Examin8\"><\/span><strong>Start Preparing Today with myCBSEguide &amp; Examin8<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Download the <strong>myCBSEguide app<\/strong> now to get access to high-quality study materials for <strong>CBSE<\/strong>, <strong>JEE<\/strong>, <strong>NEET<\/strong>, and <strong>NDA<\/strong> exams. Teachers can also leverage the <strong>Examin8 app<\/strong> to create personalized mock tests for their students and monitor their progress effectively. The <strong>CBSE Sample Papers Class 11 Chemistry<\/strong> available on myCBSEguide\u2019s app and website offer real-time exam practice, simulating actual exam conditions for students.<\/li>\n<\/ol>\n<\/div>\n<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<p style=\"text-align: center; page-break-before: always;\"><strong>Class 11 &#8211; Chemistry<br \/>\nSample Paper &#8211; 01 (2024-25)<\/strong><\/p>\n<hr \/>\n<h2 style=\"text-align: center;\"><span class=\"ez-toc-section\" id=\"Solution\"><\/span><strong>Solution <\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><b>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Section A <\/b><\/p>\n<ol style=\"padding-left: 20px;\">\n<li>(b)\n<p style=\"display: inline;\">8,17,1,8<\/p>\n<p><b>Explanation: <\/b>The balanced equation for the given equation is,<br \/>\n8CO(g) + 17H<sub>2<\/sub>(g)\u00a0<span class=\"math-tex\">{tex}\\longrightarrow{\/tex}<\/span>\u00a0C<sub>8<\/sub>H<sub>18<\/sub>(l) + 8H<sub>2<\/sub>O<\/li>\n<li>(b)\n<p style=\"display: inline;\">Positive charge of the atoms very little space<\/p>\n<p><b>Explanation: <\/b>Positive charge of the atoms very little space<\/li>\n<li>(d)\n<p style=\"display: inline;\">x &gt; y<\/p>\n<p><b>Explanation: <\/b>Here, x &gt; y. Because in equation (B) all the reactants are in their most stable states of aggregation hence have zero standard molar enthalpies of formation.<\/li>\n<li>(c)\n<p style=\"display: inline;\">16 and 15<\/p>\n<p><b>Explanation: <\/b>atomic number O has atomic number = 8 so number of electrons in <span class=\"math-tex\">{tex}O_2{\/tex}<\/span> = 16<br \/>\nwhile in <span class=\"math-tex\">{tex}O_2^+{\/tex}<\/span> there is one unit positive charge so no. of electron =15.<\/li>\n<li>(d)\n<p style=\"display: inline;\">Heat of combustion<\/p>\n<p><b>Explanation: <\/b>Combustion is an exothermic process. Hence heat of combustion has a\u00a0negative value.<\/li>\n<li>(c)\n<p style=\"display: inline;\">1s<sup>2<\/sup> 2s<sup>2<\/sup>\u00a02p<sup>6<\/sup> 3s<sup>2<\/sup>\u00a03p<sup>6<\/sup>\u00a03d<sup>5<\/sup> 4s<sup>1<\/sup><\/p>\n<p><b>Explanation: <\/b>The atomic number\u00a024 is of Cr. Due to half-filled orbital stability, it does not follow the Aufbau rule so its electronic configuration is 1s<sup>2<\/sup> 2s<sup>2<\/sup>\u00a02p<sup>6<\/sup> 3s<sup>2<\/sup>\u00a03p<sup>6<\/sup>\u00a03d<sup>5<\/sup> 4s<sup>1<\/sup>.<\/li>\n<li>(d)\n<p style=\"display: inline;\">an acid as well as an oxidant<\/p>\n<p><b>Explanation: <\/b>sulfuric acid (H<sub>2<\/sub>SO<sub>4<\/sub>) is a\u00a0strong\u00a0acid, signifying that it fully dissociates into H<sub>3<\/sub>O<sup>+<\/sup>\u00a0and HSO<sub>4<\/sub><sup>&#8211;<\/sup>\u00a0in an aqueous environment (the bisulfate ion is amphiprotic,\u00a0 but usually behaved as a weak bronsted acid given it&#8217;s feeble alkalinity).<br \/>\nNow, consider some oxidizing agents: F<sub>2<\/sub>, O<sub>2<\/sub>, Cl<sub>2<\/sub>, etc. All of these species are driven by high electronegativities\/electron affinities, a result of quantum mechanical effects that contribute to the properties of these agents. However, the H<sup>+<\/sup>\u00a0(H<sub>3<\/sub>O<sup>+<\/sup>) ion is essentially a naked proton, resulting in an\u00a0extremely\u00a0dense positive charge; this will force it&#8217;s reduction by more electropositive species by either physical gaining of electrons or the sharing of a lone pair.<br \/>\nFurthermore, the bisulfate\/sulfate ion is capable of forming entropically preferable compounds due to the presence of oxygen, another powerful oxidizing agent.<\/li>\n<li>(b)\n<p style=\"display: inline;\">free-radical mechanism<\/p>\n<p><b>Explanation: <\/b>In the presence of peroxide and light, the addition of HBr to unsymmetrical alkenes occurs contrary to Markovnikov&#8217;s rule. The chemistry follows a free-radical mechanism.\u00a0Organic reactions, which proceed by homolytic fission are called free radical or homopolar or nonpolar reactions.<\/li>\n<li>(c)\n<p style=\"display: inline;\">All of these<\/p>\n<p><b>Explanation: <\/b><\/p>\n<p>Tabacco, coal and petroleum damages DNA of our body and causes cancer.<\/li>\n<li>(a)\n<p style=\"display: inline;\">7th period and 14th group<\/p>\n<p><b>Explanation: <\/b>Elements with atomic numbers Z = (87 &#8211; 114) are placed in the 7th period.<br \/>\nThus,\u00a0 the element\u00a0with Z= 114 (Flerovium) is placed in the 7th period and 14th group of the modern periodic table.<\/li>\n<li>(b)\n<p style=\"display: inline;\">2000 K<\/p>\n<p><b>Explanation: <\/b>Gibbs free energy, <span class=\"math-tex\">{tex}\\Delta{\/tex}<\/span>G = <span class=\"math-tex\">{tex}\\Delta{\/tex}<\/span>H &#8211; T<span class=\"math-tex\">{tex}\\Delta{\/tex}<\/span>S<br \/>\nAt equilibrium <span class=\"math-tex\">{tex}\\Delta{\/tex}<\/span>G = 0;\u00a0 then T= <span class=\"math-tex\">{tex}\\frac{\\Delta H}{\\Delta S}{\/tex}<\/span> = 2000K<br \/>\nTherefore, above 2000K, the reaction will be spontaneous.<\/li>\n<li>(d)\n<p style=\"display: inline;\">NAOH + CaO<\/p>\n<p><b>Explanation: <\/b>Soda lime = NaOH + CaO<\/li>\n<li>(b)\n<p style=\"display: inline;\">Both A and R are true but R is not the correct explanation of A.<\/p>\n<p><b>Explanation: <\/b>The stability of carbonium ions is influenced by both resonance and inductive effects, allyl and benzyl carbonium ions can be stabilized by resonance but proply carbonium ion (CH<sub>3<\/sub>CH<sub>2<\/sub>CH<sub>2<\/sub><sup>+<\/sup>) has no resonating forms.<\/li>\n<li>(c)\n<p style=\"display: inline;\">A is true but R is false.<\/p>\n<p><b>Explanation: <\/b>With trans-2-butene, the product of Br<sub>2<\/sub> addition is optically inactive due to the formation of symmetric meso compounds.<br \/>\n<img decoding=\"async\" style=\"width: 250px; height: 88px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/VBhwtaT.png\" alt=\"\" data-imgur-src=\"VBhwtaT.png\" \/><\/li>\n<li>(c)\n<p style=\"display: inline;\">A is true but R is false.<\/p>\n<p><b>Explanation: <\/b>The threshold frequency is the minimum frequency required for the emission of electrons from the metal surface.<\/p>\n<ul>\n<li><strong>Comprehensive Resources<\/strong>: Access <strong>CBSE<\/strong>, <strong>JEE<\/strong>, <strong>NEET<\/strong>, and <strong>NDA<\/strong> study materials, including <strong>NCERT solutions<\/strong>, <strong>sample papers<\/strong>, and <strong>revision notes<\/strong>.<\/li>\n<li><strong>Custom Test Creation for Teachers<\/strong>: Teachers can create personalized mock tests with <strong>Examin8<\/strong> to align with their curriculum and better support student learning.<\/li>\n<li><strong>User-Friendly Apps<\/strong>: Both <strong>myCBSEguide<\/strong> and <strong>Examin8<\/strong> are available for <strong>iOS<\/strong> and <strong>Android<\/strong>, providing students and teachers with easy access to study materials and tests anytime, anywhere.<\/li>\n<li><strong>Boost Exam Performance<\/strong>: With the right tools and resources, both students and teachers can achieve better results and improved learning outcomes.(d)\n<p style=\"display: inline;\">A is false but R is true.<\/p>\n<p><b>Explanation: <\/b>Gases do deviate from ideal behaviour but under the same conditions of pressure, temperature, etc., they combine in a simple ratio of volume fairly accurately.<\/p>\n<h3><span class=\"ez-toc-section\" id=\"Download_the_myCBSEguide_App_for_Complete_Exam_Preparation\"><\/span><strong>Download the myCBSEguide App for Complete Exam Preparation<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>For comprehensive preparation and practice for <strong>CBSE<\/strong>, <strong>JEE (Main)<\/strong>, <strong>NEET-UG<\/strong>, and <strong>NDA<\/strong> exams, download the <strong><a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide\">myCBSEguide<\/a> App and <a href=\"https:\/\/mycbseguide.com\/\">myCBSEGuide<\/a> Website<\/strong>. This all-in-one learning app offers everything you need to excel:<\/p>\n<ul>\n<li><strong>Complete Study Material<\/strong> for all <strong>CBSE<\/strong> subjects.<\/li>\n<li><strong>NCERT solutions<\/strong> and <strong>chapter-wise practice tests<\/strong> to strengthen your concepts.<\/li>\n<li><strong>Sample papers<\/strong> and <strong>CBSE guess papers<\/strong> to help you focus on high-yield topics.<\/li>\n<li><strong>Quick revision notes<\/strong> for last-minute preparation and effective study sessions.<\/li>\n<\/ul>\n<p>With these resources, the <strong>myCBSEguide app<\/strong> ensures that you&#8217;re well-prepared and confident to tackle your exams.<\/p>\n<p><strong>Create Custom Mock Tests with Examin8.com App<\/strong><\/p>\n<p>For <strong>teachers<\/strong>, the\u00a0<strong><a href=\"https:\/\/play.google.com\/store\/apps\/details?id=com.examin8.exam\">Examin8<\/a> App and <a href=\"https:\/\/examin8.com\/\">Examin8<\/a> Website<\/strong>\u00a0is an excellent tool to design custom <strong>mock tests<\/strong> and practice papers. To better prepare for board exams, practicing <strong>CBSE Sample Papers Class 11 Chemistry<\/strong> regularly can help students track their progress and identify areas that need improvement. Key features include:<\/p>\n<ul>\n<li><strong>Custom Branding<\/strong>: Add your <strong>name<\/strong> and <strong>logo<\/strong> to the tests for a professional and personalized touch.<\/li>\n<li><strong>Flexible Question Formats<\/strong>: Choose from multiple question formats, making it easy to design assessments that align with your teaching.<\/li>\n<li><strong>Seamless Test Distribution<\/strong>: Distribute your tests to students effortlessly and track their performance in real-time.<\/li>\n<\/ul>\n<p>The <strong>Examin8 app<\/strong> is a game-changer for educators looking to provide tailored assessments and better track student progress.<\/li>\n<\/ul>\n<\/li>\n<li style=\"text-align: center; display: block;\"><b>Section B <\/b><\/li>\n<li><span class=\"math-tex\">{tex}\\mathrm{A}_{2} \\mathrm{X}_{3} \\rightarrow 2 \\mathrm{A}^{3 +}+3 \\mathrm{X}^{2-}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}K_{\\mathrm{sp}}=\\left[\\mathrm{A}^{3+}\\right]^{2}\\left[\\mathrm{X}^{2-}\\right]^{3}=1.1 \\times 10^{-23}{\/tex}<\/span><br \/>\nIf S = solubility of A<sub>2<\/sub>X<sub>3<\/sub>, then<br \/>\n<span class=\"math-tex\">{tex}\\left[\\mathrm{A}^{3+}\\right]=2 \\mathrm{S} ;\\left[\\mathrm{X}^{2}\\right]=3 \\mathrm{S}{\/tex}<\/span><br \/>\ntherefore, K<sub>sp<\/sub> = (2S)<sup>2<\/sup>(3S)<sup>3<\/sup> = 108S<sup>5<\/sup><br \/>\n= 1.1 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span> 10<sup>\u201323<\/sup><br \/>\nthus, S<sup>5<\/sup> = 1 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span> 10<sup>\u201325<\/sup><br \/>\nS = 1.0 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span> 10<sup>\u20135<\/sup> mol\/L<\/li>\n<li>The electronic configuration of helium (He) is 1s<sup><sub>2<\/sub><\/sup>. It\u00a0should be kept in s-block as the last electron enters in &#8216;s&#8217; orbital. Due to the presence of completely filled\u00a0valence shell\u00a0(1s<sup><sub>2<\/sub><\/sup> ) and similarities in properties with noble gases, it is placed in p-block i.e. group 18.<\/li>\n<li>According to the chemical equation,<br \/>\nCH<sub>4<\/sub> (g) + 2O<sub>2<\/sub>(g) <span class=\"math-tex\">{tex}\\to{\/tex}<\/span>\u00a0CO<sub>2<\/sub>(g) + 2H<sub>2<\/sub>O (g)<br \/>\n44g CO<sub>2<\/sub> (g) is obtained from 16 g CH<sub>4<\/sub> (g).<br \/>\n[<span class=\"math-tex\">{tex}\\therefore{\/tex}<\/span>\u00a01 mol CO<sub>2<\/sub>(g) is obtained from 1 mol of CH<sub>4<\/sub>(g)]\nNumber of moles of CO<sub>2<\/sub> (g)<br \/>\n= 22 g CO<sub>2<\/sub> (g) <span class=\"math-tex\">{tex}\\times{\/tex}<\/span><span class=\"math-tex\">{tex}\\frac{1 \\mathrm{mol} \\mathrm{CO}_{2}(\\mathrm{g})}{44 \\mathrm{gCO}_{2}(\\mathrm{g})}{\/tex}<\/span><br \/>\n= 0.5 mol CO<sub>2<\/sub>(g)<br \/>\nHence, 0.5 mol CO<sub>2<\/sub> (g) would be obtained from 0.5 mol CH<sub>4 <\/sub>(g) or 0.5 mol of CH<sub>4<\/sub> (g) would be required to produce 22 g CO<sub>2<\/sub> (g).<\/li>\n<li style=\"clear: both;\">For the given compound, the number of <span class=\"math-tex\">{tex}\\pi{\/tex}<\/span>-electrons is 8.<br \/>\nBy Huckel\u2019s rule,<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a04<i>n<\/i>\u00a0+ 2 = 8<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a04<i>n<\/i>\u00a0= 6<br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow{\/tex}<\/span>\u00a0n = 3\/2<br \/>\nFor a compound to be aromatic, the value of n must be an integer (n = 0, 1, 2\u2026).<br \/>\nThis is not true for the given compound as it is a fraction. Hence, it is not aromatic in nature.<\/p>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p>Benzene,\u00a0when undergoes a process of a complete oxidation reaction by V<sub>2<\/sub>O<sub>5<\/sub>, gives Maleic Anhydride.<br \/>\n<img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/F1io2Qd.png\" data-imgur-src=\"F1io2Qd.png\" \/><\/li>\n<li>\u00a0Given, Mass of electron m<sub>e<\/sub>\u00a0= 9.1 \u00d7\u00a010<sup>-31<\/sup>\u00a0kg and\u00a0h = 6.6\u00a0\u00d7\u00a010<sup>-34<\/sup>\u00a0kg m<sup>2<\/sup>s<sup>-1<\/sup>Uncertainty in speed,\u00a0<span class=\"math-tex\">{tex}\\Delta v=600 \\times \\frac{0.005}{100}=0.03 \\mathrm{ms}^{-1}{\/tex}<\/span>Using Heisenberg\u2019s uncertainty principle, <span class=\"math-tex\">{tex}\\Delta x=\\frac{6.6 \\times 10^{-34}}{4 \\times 3.14 \\times 9.1 \\times 10^{-31} \\times 0.03}{\/tex}<\/span><span class=\"math-tex\">{tex}=1.92 \\times 10^{-3} \\mathrm{m}{\/tex}<\/span>.<br \/>\n.<\/li>\n<li style=\"text-align: center; display: block;\"><b>Section C <\/b>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>The resonating structures of O<sub>3<\/sub>\u00a0are shown below:<br \/>\n<img decoding=\"async\" style=\"width: 300px; height: 60px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/quesbank\/12\/sci\/ch07\/img02.png\" \/><\/li>\n<li>Nitrate ion, <span class=\"math-tex\">{tex}NO^-_3{\/tex}<\/span><br \/>\n<img decoding=\"async\" style=\"width: 200px; height: 53px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1683547006-nrmabw.jpg\" \/><br \/>\n<img decoding=\"async\" style=\"width: 100px; height: 72px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/question_images\/1683547042-z8gscy.jpg\" \/><\/li>\n<\/ol>\n<\/li>\n<li>Answer:\n<ol class=\"inner-group-question-ol\">\n<li style=\"clear: both;\">We have q = -w = p<sub>ex<\/sub> (8) = 8 litre-atm<\/li>\n<li style=\"clear: both;\">The <strong>Third Law of Thermodynamics <\/strong>states, \u201cThe entropy of a perfect crystal is zero when the temperature of the crystal is equal to <strong>absolute zero (0 K).<\/strong><\/li>\n<li style=\"clear: both;\">For an isolated system at constant volume, there is no transfer of energy in the form of\u00a0heat or work.<br \/>\nSo,\u00a0<span class=\"math-tex\">{tex}\\Delta U = q + W{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\Delta U = 0+0{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\Rightarrow \\Delta U =0{\/tex}<\/span><\/li>\n<\/ol>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>A substance has perfectly ordered arrangement of its constituent particles only at absolute zero. When the element from itself. This means no heat change.<br \/>\nThus\u00a0<span class=\"math-tex\">{tex}\\Delta{\/tex}<\/span><sub>f<\/sub>H = 0<\/p>\n<ol style=\"list-style-type: lower-alpha;\" start=\"1\">\n<li>If both\u00a0<span class=\"math-tex\">{tex}\\Delta{\/tex}<\/span>H and\u00a0<span class=\"math-tex\">{tex}\\Delta{\/tex}<\/span>S are positive <span class=\"math-tex\">{tex}\\Delta{\/tex}<\/span>G can be -ve only in magnitude. Thus the temperature should be high.<\/li>\n<li>If both <span class=\"math-tex\">{tex}\\Delta{\/tex}<\/span>H and <span class=\"math-tex\">{tex}\\Delta{\/tex}<\/span>S are negative <span class=\"math-tex\">{tex}\\Delta{\/tex}<\/span>G can be negative only T<span class=\"math-tex\">{tex}\\Delta{\/tex}<\/span>S &lt;\u00a0<span class=\"math-tex\">{tex}\\Delta{\/tex}<\/span>H is magnitude. Thus the value of T should be low.<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>At anode: Zn <span class=\"math-tex\">{tex}\\rightarrow{\/tex}<\/span> Zn<sup>2+<\/sup> + 2e<sup>&#8211; <\/sup>&#8230;(i)<br \/>\nAt cathode: Br<sub>2<\/sub> + 2e<sup>&#8211;<\/sup> <span class=\"math-tex\">{tex}\\rightarrow{\/tex}<\/span> 2Br<sup><span style=\"font-size: 10.8333px;\">&#8211;<\/span><\/sup> &#8230;.(ii)<br \/>\non adding (i) and (ii)<br \/>\nZn + Br<sub>2<\/sub> <span class=\"math-tex\">{tex}\\rightarrow{\/tex}<\/span> Zn<sup>2+<\/sup> + 2 Br<sup>&#8211;<\/sup><\/li>\n<li>At anode: [Cr <span class=\"math-tex\">{tex}\\rightarrow{\/tex}<\/span> Cr<sup>3<\/sup> + 3e<sup>&#8211; <\/sup> ] <span class=\"math-tex\">{tex}\\times{\/tex}<\/span> 2 &#8230;(iii)<br \/>\nAt cathode: [I2 + 2e<span style=\"font-size: 10.8333px;\"><sup>&#8211;<\/sup> <span class=\"math-tex\">{tex}\\rightarrow{\/tex}<\/span><\/span> 2I<sup>&#8211;<\/sup>] <span class=\"math-tex\">{tex}\\times{\/tex}<\/span> 3 &#8230;(iv)<br \/>\non adding (iii) and (iv)<br \/>\n2Cr + 3I<sub>2<\/sub> <span class=\"math-tex\">{tex}\\rightarrow{\/tex}<\/span> 6I<sup>&#8211;<\/sup> + 2Cr<sup>3+<\/sup><\/li>\n<li>At anode: H<sub>2<\/sub>(g) <span class=\"math-tex\">{tex}\\rightarrow{\/tex}<\/span> 2H<sup>+<\/sup> (aq) + 2e<sup>&#8211;<\/sup> &#8230;(v)<br \/>\nAt cathode: Cu<sup>2+<\/sup> (aq) + 2e<sup>&#8211;<\/sup> <span class=\"math-tex\">{tex}\\rightarrow{\/tex}<\/span> Cu(s) &#8230;(vi)<br \/>\non adding (v) and (vi)<br \/>\nH<sub>2<\/sub>(g) + Cu<sup>2+<\/sup> (aq) <span class=\"math-tex\">{tex}\\rightarrow{\/tex}<\/span> Cu(s) + 2H<sup>+<\/sup> (aq)<br \/>\nCathode will be positive terminal in each cell.<\/li>\n<\/ol>\n<\/li>\n<li>v = (3.29 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span>10<sup>15 <\/sup>Hz)\u00a0<span class=\"math-tex\">{tex}\\left(\\frac{1}{3^{2}}-\\frac{1}{n^{2}}\\right){\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\lambda{\/tex}<\/span>\u00a0= 1285 nm = 1285\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a010<sup>-9<\/sup> m = 1.285\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>10<sup>-6<\/sup> m<br \/>\nv =\u00a0<span class=\"math-tex\">{tex}\\frac{c}{\\lambda}=\\frac{\\left(3 \\times 10^{8} \\mathrm{ms}^{-1}\\right)}{\\left(1.285 \\times 10^{-6} \\mathrm{m}\\right)}{\/tex}<\/span>\u00a0= 2.3346 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a010<sup>14<\/sup>s<sup>-1<\/sup><br \/>\n2.3346\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>10<sup>14<\/sup> = 3.29\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>10<sup>15\u00a0<\/sup><span class=\"math-tex\">{tex}\\left[\\frac{1}{3^{2}}-\\frac{1}{\\mathrm{n}^{2}}\\right]{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\frac{2.3346}{32.9}=\\frac{1}{3^{2}}-\\frac{1}{\\mathrm{n}^{2}}{\/tex}<\/span>or 0.71 =\u00a0<span class=\"math-tex\">{tex}\\frac{1}{9}-\\frac{1}{n^{2}}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}\\frac{1}{n^{2}}=\\frac{1}{9}{\/tex}<\/span>&#8211; 0.071 = 0.111 &#8211; 0.071 = 0.04<br \/>\nn<sup>2<\/sup> =\u00a0<span class=\"math-tex\">{tex}\\frac{1}{0.04}{\/tex}<\/span>= 25 or n = 5<br \/>\nPaschen series lies in infrared region of the spectrum.<\/p>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>C\u00a0has the highest first ionization enthalpy.<\/li>\n<li>C\u00a0has the most negative electron gain enthalpy.<\/li>\n<li>AI\u00a0\u00a0has the largest atomic radius.<\/li>\n<li>AI\u00a0\u00a0has the most metallic character.<\/li>\n<\/ol>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li>Molecular mass of H<sub>2<\/sub>O = 2\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>H + 1\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a0O = 2\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a01.0079u + 1\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a016.00u = 18.0158 u<\/li>\n<li>Molecular mass of CO<sub>2<\/sub> =1<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a0C + 2 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a0O = 1 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span>12.01 u + 2<span class=\"math-tex\">{tex}\\times{\/tex}<\/span> 16.22 u = 44.01 u<\/li>\n<li>Molecular mass of CH<sub>4<\/sub>\u00a0= 1\u00a0<span class=\"math-tex\">{tex}\\times{\/tex}<\/span>\u00a0C+ 4 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span> H = 1 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span>12.01 u + 4 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span> 1.0079 u = 16.0416 u<\/li>\n<\/ol>\n<\/li>\n<li style=\"text-align: center; display: block;\"><b>Section D <\/b>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li style=\"text-align: left;\">Fractional crystallizationis used to separate two compounds with different solubilities in a solvent.\n<ul>\n<li style=\"text-align: left;\">volatile liquids from nonvolatile impurities.<\/li>\n<li style=\"text-align: left;\">the liquids having sufficient difference in their boiling points.<\/li>\n<\/ul>\n<\/li>\n<li style=\"text-align: left;\">Aniline is separated from aniline water mixture by steam distillation as one of the substances in the mixture is water and the other, a water insoluble substance.<br \/>\n<strong>OR<\/strong><br \/>\nChloroform and aniline are easily separated by the technique of distillation because chloroform and aniline have sufficient difference in their boiling points.<\/li>\n<\/ol>\n<ol style=\"list-style-type: lower-roman;\" start=\"1\">\n<li style=\"text-align: left;\">The\u00a0increasing order\u00a0of negative stabilities of CN, CN<sup>+<\/sup> and CN<sup>&#8211;<\/sup>\u00a0 is CN<sup>+<\/sup> &gt; CN &gt; CN<sup>&#8211;<\/sup>.<\/li>\n<li style=\"text-align: left;\">The molecular orbital theory is preferred over valence bond theory\u00a0because molecular orbital theory explains the magnetic nature of the molecule.<\/li>\n<li style=\"text-align: left;\">In ethyne, hydrogen atoms are connected to sp hybridized carbon atoms, but in ethene, they are attached to sp<sup>2<\/sup> hybridized carbon atoms and in ethane, they are attached to sp<sup>3<\/sup> hybridized carbons.<br \/>\n<strong>OR<\/strong><br \/>\nThe given statement is not correct because the bonding molecular orbital is lowered by a lesser amount of energy than the amount by which antibonding molecular orbital is raised.<\/li>\n<\/ol>\n<\/li>\n<li style=\"text-align: center; display: block;\"><b>Section E <\/b><\/li>\n<li>Attempt any five of the following:\n<ol class=\"inner-group-question-ol\">\n<li style=\"clear: both;\">Hydrocarbons are categorized into three categories according to the carbon-carbon bond that exists between them:\n<ol style=\"list-style-type: lower-alpha;\" start=\"1\">\n<li>Saturated hydrocarbon (In which carbon-carbon single bond are present)<\/li>\n<li>Unsaturated hydrocarbon (In which carbon-carbon double and triple bonds are present)<\/li>\n<li>Aromatic hydrocarbon\u00a0(In which alternate single and double bond and (4n+2)<span class=\"math-tex\">{tex}\\pi{\/tex}<\/span>\u00a0electrons are present)<\/li>\n<\/ol>\n<\/li>\n<li style=\"clear: both;\">The triple bonds of\u00a0alkynes, because of its high electron density, are easily attacked by electrophiles, but\u00a0less reactive than alkenes\u00a0due\u00a0to\u00a0the compact C-C electron cloud. The three-membered ring bromonium ion formed from the alkyne (A) has a full double bond causing it to be more stained and less stable than the one from the alkene (B),<img decoding=\"async\" style=\"width: 296px; height: 107px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/AjxZMvj.png\" alt=\"\" data-imgur-src=\"AjxZMvj.png\" \/><br \/>\nAlso, the carbon\u2019s of (A) that are part of the bromonium ion has more s-character than (B), further making (A) less stable than (B).<\/li>\n<li style=\"clear: both;\">SO<sub>3<\/sub><\/li>\n<li style=\"clear: both;\">Alkenes are commonly known as olefins because the lower members form oily products on treatment with chlorine or bromine.<\/li>\n<li style=\"clear: both;\">Iso-propylmagnesium bromide, (CH<sub>3<\/sub>)<sub>2<\/sub>CHMgBr,<br \/>\n(CH<sub>3<\/sub>)<sub>2<\/sub>CHMgBr +\u00a0H<sub>2<\/sub>O\u00a0<span class=\"math-tex\">{tex}\\longrightarrow{\/tex}<\/span>\u00a0CH<sub>3<\/sub>CH<sub>2<\/sub>CH<sub>3<\/sub> + Mg(OH)Br<\/li>\n<li style=\"clear: both;\">2-methylbutane<\/li>\n<li style=\"clear: both;\">In trans-but-2-ene, the dipole moments of the two C\u2014CH<sub>3<\/sub> bonds are equal and opposite and therefore, they cancel out each other.<br \/>\nHence, trans-2-butene is non-polar.<br \/>\n<img decoding=\"async\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/imgur\/t0q5dXL.png\" alt=\"\" data-imgur-src=\"t0q5dXL.png\" \/><\/li>\n<\/ol>\n<\/li>\n<li style=\"clear: both;\">For the reaction,<br \/>\nlet \u2018x\u2019 be the decrease in pressure of CO<sub>2<\/sub>, then<br \/>\nCO<sub>2<\/sub>(g) + C(s)\u00a0<span class=\"math-tex\">{tex} \\rightleftharpoons {\/tex}<\/span>\u00a02CO(g)<br \/>\nInitial<br \/>\npressure: 0.48 bar 0<br \/>\nAt equilibrium:\u00a0(0.48 \u2013 x)bar 2x bar<br \/>\n<span class=\"math-tex\">{tex}K_{p}=\\frac{p_{C O}^{2}}{p_{C O_{2}}}{\/tex}<\/span><br \/>\nK<sub>p<\/sub> = (2x)<sup>2<\/sup>\/(0.48 \u2013 x) = 3<br \/>\n4x<sup>2<\/sup> = 3(0.48 \u2013 x)<br \/>\n4x<sup>2<\/sup> = 1.44 \u2013 x<br \/>\n4x<sup>2<\/sup> + 3x \u2013 1.44 = 0<br \/>\na = 4, b = 3, c = \u20131.44<br \/>\nx =\u00a0<span class=\"math-tex\">{tex}\\frac{(-b \\pm \\sqrt{b^{2}-4 a c})}{2 a}{\/tex}<\/span><br \/>\n<span class=\"math-tex\">{tex}x = {-3 \\pm \\sqrt{3^2-4(4)(-1.44)} \\over 2(4)}{\/tex}<\/span><br \/>\n= (\u20133 <span class=\"math-tex\">{tex}\\pm{\/tex}<\/span> 5.66)\/8<br \/>\n= (\u20133 + 5.66)\/8 (as value of x cannot be negative hence we neglect that value)<br \/>\nx = 2.66\/8 = 0.33<br \/>\nThe equilibrium partial pressures are,<br \/>\n<span class=\"math-tex\">{tex}P_{CO_2}{\/tex}<\/span> = 2x = 2 <span class=\"math-tex\">{tex}\\times{\/tex}<\/span> 0.33 = 0.66 bar<br \/>\n<span class=\"math-tex\">{tex}P_{CO_2}{\/tex}<\/span> = 0.48 \u2013 x = 0.48 \u2013 0.33 = 0.15 bar<\/p>\n<p style=\"text-align: center;\"><b>OR <\/b><\/p>\n<p><img decoding=\"async\" style=\"height: 30px; width: 200px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/quesbank\/11\/chem\/ch-07\/image647.png\" \/><br \/>\n<span class=\"math-tex\">{tex}{Q_p} = \\frac{{pc{o_2}}}{{pco}} = \\frac{{(0.8\\;atm)}}{{(1.4\\;atm)}} = 0.571{\/tex}<\/span><br \/>\nSince Q<sub>p<\/sub> &gt; k<sub>p<\/sub> (0.265), this means that the reaction will move in the backward direction to attain the equilibrium. Therefore, the partial pressure of CO<sub>2<\/sub> will decrease while that of CO will increase so that the equilibrium may be attained again. Let p atm be the decrease in the partial pressure of CO<sub>2<\/sub>. Therefore, the partial pressure of CO will increase by the same magnitude i.e. p atm.<br \/>\n<span class=\"math-tex\">{tex}P_{CO_2}{\/tex}<\/span>\u00a0= (0.8 &#8211; p) atm; <span class=\"math-tex\">{tex}P_{CO}{\/tex}<\/span>(g) = (1.4 + p) atm<br \/>\nAt equilibrium <span class=\"math-tex\">{tex}{K_p} = \\frac{{p_{co_2}}}{{p_{co}}} = \\frac{{(0.8 &#8211; p)\\;atm}}{{(1.4 + p)\\;atm}} = \\frac{{(0.8 &#8211; p)}}{{(1.4 + p)}}{\/tex}<\/span><br \/>\nor <span class=\"math-tex\">{tex}0.265 = \\frac{{(0.8 &#8211; p)}}{{(1.4 + p)}}{\/tex}<\/span><br \/>\n0.371 + 0.265 p = 0.8 &#8211; p or 1.265 p = 0.8 &#8211; 0.371 = 0.429<br \/>\np = 0.429 \/ 1.265 = 0.339 atm<br \/>\nThe equilibrium partial pressure of CO is (<span class=\"math-tex\">{tex}P_{CO}{\/tex}<\/span>)<sub>eq<\/sub> = (1.4 + 0.339) = 1.739 atm<br \/>\nThe equilibrium partial pressure of CO<sub>2\u00a0<\/sub>is (<span class=\"math-tex\">{tex}P_{CO_2}{\/tex}<\/span>)eq = (0.8 &#8211; 0.339) = 0.461 atm<\/li>\n<li>Answer:\n<ol class=\"inner-group-question-ol\">\n<li class=\"question-list\" style=\"clear: both;\">\n<div style=\"margin-left: 30px;\">\n<ol style=\"list-style: lower-roman !important;\" type=\"i\">\n<li style=\"clear: both;\">Chromatography is based on the principle of different adsorption.<\/li>\n<li style=\"clear: both;\">Monohydric alcohols are the compounds derived from an alkane by replacing one H by &#8211; OH group.<br \/>\nExample:<br \/>\n<span class=\"math-tex\">{tex}\\mathop {{\\text{C}}{{\\text{H}}_4}}\\limits_{{\\text{Methane}}} \\xrightarrow{ replacing\\ H \\ with \\ OH }\\mathop {{\\text{C}}{{\\text{H}}_{\\text{3}}} &#8211; {\\text{OH}}}\\limits_{{\\text{Methanol}}} {\/tex}<\/span><br \/>\nTherefore, the general molecular formula of saturated monohydric alcohols is\u00a0C<sub>n<\/sub>H<sub>2n+1<\/sub>OH.<\/li>\n<\/ol>\n<\/div>\n<\/li>\n<li class=\"question-list nostyle\" style=\"clear: both;\">\n<div class=\"col-12\" style=\"text-align: center;\"><b>OR<\/b><\/div>\n<div style=\"margin-left: 30px;\">\n<ol style=\"list-style: lower-roman !important;\" type=\"i\">\n<li style=\"clear: both;\">This method is based upon the fact that nitrogenous compound is heated with copper oxide in an atmosphere of carbondioxide yield free nitrogen.<span class=\"math-tex\">{tex}\\mathrm{C}_{x} \\mathrm{H}_{y} \\mathrm{N}_{z}+\\mathrm{CuO} \\longrightarrow x \\mathrm{CO}_{2}+\\frac{y}{2} \\mathrm{H}_{2} \\mathrm{O}+\\frac{z}{2} \\mathrm{N}_{2}+(\\mathrm{Cu}){\/tex}<\/span><\/li>\n<li style=\"clear: both;\"><strong>Hyperconjugation:<\/strong> The relative stability of various classes of carbonium ions may be explained by the number of no-bond resonance structures that can be written for them. Such structures are arrived by shifting the bonding electrons from an adjacent C &#8211; H bond to the electron &#8211; deficient carbon. In this way, the positive charge originally on carbon is dispersed to the hydrogen. This manner of electron release by assuming no-bond character in the adjacent C &#8211; H bond is called Hyper conjugation or No-bond Resonance. The greater the hyperconjugation, the greater will be the stability of the compound. The increasing order of stability can be shown as:<br \/>\n<img decoding=\"async\" style=\"height: 30px; width: 176px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/quesbank\/11\/chem\/ch-12\/image174.png\" \/><img decoding=\"async\" style=\"height: 57px; width: 164px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/quesbank\/11\/chem\/ch-12\/image175.png\" \/><img decoding=\"async\" style=\"height: 58px; width: 129px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/quesbank\/11\/chem\/ch-12\/image176.png\" \/><\/p>\n<h3><span class=\"ez-toc-section\" id=\"Download_myCBSEguide_App_for_Comprehensive_Exam_Preparation\"><\/span><strong>Download myCBSEguide App for Comprehensive Exam Preparation<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Looking to excel in your <strong>CBSE<\/strong>, <strong>JEE (Main)<\/strong>, <strong>NEET-UG<\/strong>, or <strong>NDA<\/strong> exams? The <strong><a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide\">myCBSEguide<\/a> App and <a href=\"https:\/\/mycbseguide.com\/\">myCBSEGuide<\/a> Website<\/strong>\u00a0is your ultimate study companion. With the app, students can access:<\/p>\n<ul>\n<li><strong>Complete Study Material<\/strong>: Get NCERT solutions, chapter-wise tests, and practice papers.<\/li>\n<li><strong>JEE, NEET, and NDA Resources<\/strong>: In addition to CBSE, access comprehensive study material for competitive exams like <strong>JEE (Main)<\/strong>, <strong>NEET-UG<\/strong>, and <strong>NDA<\/strong>.<\/li>\n<li><strong>Sample Papers &amp; Guess Papers<\/strong>: Practice with CBSE-specific sample papers and important question papers to boost exam performance.<\/li>\n<li><strong>Quick Revision Notes<\/strong>: Perfect for last-minute revision, our concise notes help you focus on key concepts and topics.<\/li>\n<\/ul>\n<p>With <strong>myCBSEguide<\/strong>, students can stay organized, track progress, and confidently prepare for their exams. The <strong>CBSE Sample Papers Class 11 Chemistry<\/strong> provided by myCBSEguide are designed according to the latest syllabus and exam guidelines to give students an accurate practice experience.<\/p>\n<p><strong>Create Custom Mock Tests with Examin8 App<\/strong><\/p>\n<p>For <strong>teachers<\/strong>, the\u00a0<strong><a href=\"https:\/\/play.google.com\/store\/apps\/details?id=com.examin8.exam\">Examin8<\/a> App and <a href=\"https:\/\/examin8.com\/\">Examin8<\/a> Website<\/strong>\u00a0offers a powerful platform to create <strong>custom mock tests<\/strong> and practice papers. Key features include:<\/p>\n<ul>\n<li><strong>Custom Branding<\/strong>: Teachers can add their own <strong>name<\/strong> and <strong>logo<\/strong> to tests, ensuring a personalized touch.<\/li>\n<li><strong>Flexible Question Formats<\/strong>: Choose from a variety of question types and formats, suitable for any curriculum or exam style.<\/li>\n<li><strong>Easy Distribution<\/strong>: Create and distribute tests easily to students, enabling effective monitoring and assessment.<\/li>\n<\/ul>\n<p>The <strong>Examin8 app<\/strong> is ideal for teachers who want to provide customized, high-quality assessments that align with their teaching objectives.<\/p>\n<p><strong>Why Choose myCBSEguide &amp; Examin8?<\/strong><\/p>\n<ul>\n<li><strong>Complete Study Resources<\/strong>: Access study material for <strong>CBSE<\/strong>, <strong>JEE<\/strong>, <strong>NEET<\/strong>, and <strong>NDA<\/strong>, including sample papers, NCERT solutions, and revision notes.<\/li>\n<li><strong>Custom Mock Test Creation<\/strong>: Teachers can design personalized mock tests with the <strong>Examin8<\/strong> app, perfect for effective student assessments.<\/li>\n<li><strong>Accessible Anytime, Anywhere<\/strong>: Both the <strong>myCBSEguide<\/strong> and <strong>Examin8<\/strong> apps are available on <strong>iOS<\/strong> and <strong>Android<\/strong>, making it easy to study or create tests on the go.<\/li>\n<li><strong>Boost Exam Performance<\/strong>: With tailored study materials and custom mock test tools, both students and teachers can significantly improve their exam preparation and results.<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n<\/div>\n<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<p style=\"text-align: center;\"><strong><a class=\"button\" href=\"https:\/\/mycbseguide.com\/dashboard\/category\/1356\/type\/2\">Download Chemistry Sample Papers\u00a0as PDF<\/a><\/strong><\/p>\n<p>&nbsp;<\/p>\n<h2><span class=\"ez-toc-section\" id=\"Marking_Scheme_for_the_Class_11_exam\"><\/span><strong>Marking Scheme for the Class 11 exam<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<table class=\"mobile\" border=\"1\" cellspacing=\"0\" cellpadding=\"3\">\n<thead>\n<tr>\n<th>Subject<\/th>\n<th>Board Marks<\/th>\n<th>Practical or internal Marks<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>English<\/td>\n<td>80 Marks<\/td>\n<td>20 Marks<\/td>\n<\/tr>\n<tr>\n<td>Hindi<\/td>\n<td>80 Marks<\/td>\n<td>20 Marks<\/td>\n<\/tr>\n<tr>\n<td>Mathematics<\/td>\n<td>80 Marks<\/td>\n<td>20 Marks<\/td>\n<\/tr>\n<tr>\n<td>Chemistry<\/td>\n<td>70 Marks<\/td>\n<td>30 Marks<\/td>\n<\/tr>\n<tr>\n<td>Physics<\/td>\n<td>70 Marks<\/td>\n<td>30 Marks<\/td>\n<\/tr>\n<tr>\n<td>Biology<\/td>\n<td>70 Marks<\/td>\n<td>30 Marks<\/td>\n<\/tr>\n<tr>\n<td>Computer Science<\/td>\n<td>70 Marks<\/td>\n<td>30 Marks<\/td>\n<\/tr>\n<tr>\n<td>Informatics Practices<\/td>\n<td>70 Marks<\/td>\n<td>30 Marks<\/td>\n<\/tr>\n<tr>\n<td>Chemistry<\/td>\n<td>80 Marks<\/td>\n<td>20 Marks<\/td>\n<\/tr>\n<tr>\n<td>Business Studies<\/td>\n<td>80 Marks<\/td>\n<td>30 Marks<\/td>\n<\/tr>\n<tr>\n<td>Economics<\/td>\n<td>80 Marks<\/td>\n<td>20 Marks<\/td>\n<\/tr>\n<tr>\n<td>History<\/td>\n<td>80 Marks<\/td>\n<td>20 Marks<\/td>\n<\/tr>\n<tr>\n<td>Political Science<\/td>\n<td>80 Marks<\/td>\n<td>20 Marks<\/td>\n<\/tr>\n<tr>\n<td>Geography<\/td>\n<td>70 Marks<\/td>\n<td>30 Marks<\/td>\n<\/tr>\n<tr>\n<td>Sociology<\/td>\n<td>80 Marks<\/td>\n<td>20 Marks<\/td>\n<\/tr>\n<tr>\n<td>Physical Education<\/td>\n<td>70 Marks<\/td>\n<td>30 Marks<\/td>\n<\/tr>\n<tr>\n<td>Home Science<\/td>\n<td>70 Marks<\/td>\n<td>30 Marks<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h2><span class=\"ez-toc-section\" id=\"CBSE_Sample_Papers_for_Class_11\"><\/span><strong>CBSE\u00a0Sample Papers\u00a0for Class 11<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<ul>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-11-physics\/1340\/\"><strong>Physics<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-11-chemistry\/1356\/\"><strong>Chemistry<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-11-mathematics\/1371\/\"><strong>Mathematics<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-11-biology\/1388\/\"><strong>Biology<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-11-accountancy\/1411\/\"><strong>Accountancy<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-11-economics\/1423\/\"><strong>Economics<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-11-business-studies\/1740\/\"><strong>Business Studies<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-11-computer-science\/1852\/\"><strong>Computer Science<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-11-informatics-practices\/1874\/\"><strong>Informatics Practices<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-11-english-core\/1856\/\"><strong>English Core<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-11-%E0%A4%B9%E0%A4%BF%E0%A4%82%E0%A4%A6%E0%A5%80-%E0%A4%95%E0%A5%8B%E0%A4%B0\/1866\/\"><strong>Hindi Core<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-11-%E0%A4%B9%E0%A4%BF%E0%A4%82%E0%A4%A6%E0%A5%80-%E0%A4%90%E0%A4%9A%E0%A5%8D%E0%A4%9B%E0%A4%BF%E0%A4%95\/1868\/\"><strong>Hindi Elective<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-11-history\/1870\/\"><strong>History<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-11-political-science\/1880\/\"><strong>Political Science<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-11-geography\/1864\/\"><strong>Geography<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-11-sociology\/1882\/\"><strong>Sociology<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-11-physical-education\/1878\/\"><strong>Physical Education<\/strong><\/a><\/li>\n<li><strong><a href=\"https:\/\/mycbseguide.com\/course\/cbse-class-11\/1339\/\">Other Subjects<\/a><\/strong><\/li>\n<li><\/li>\n<\/ul>\n<h2><span class=\"ez-toc-section\" id=\"CBSE_Class_11_Chemistry_Model_Papers\"><\/span>CBSE Class 11 Chemistry Model Papers<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>Looking for <strong>Class 11 sample papers<\/strong> for <strong>Physics<\/strong>, <strong>Chemistry<\/strong>, <strong>Biology<\/strong>, <strong>History<\/strong>, <strong>Political Science<\/strong>, <strong>Economics<\/strong>, <strong>Geography<\/strong>, <strong>Computer Science<\/strong>, <strong>Home Science<\/strong>, <strong>Accountancy<\/strong>, and <strong>Business Studies<\/strong>? Get all your exam preparation materials on the <strong>myCBSEguide<\/strong> app and website. By practicing with <strong>CBSE Sample Papers Class 11 Chemistry<\/strong>, students can improve their problem-solving skills and boost their confidence for the upcoming board exams.<\/p>\n<p><strong>myCBSEguide<\/strong> provides comprehensive study resources including:<\/p>\n<ul>\n<li><strong>Sample Papers with Solutions<\/strong>: Practice with solved sample papers to understand exam patterns and improve your performance.<\/li>\n<li><strong>Chapter-wise Test Papers<\/strong>: Strengthen your understanding of each subject with test papers based on specific chapters.<\/li>\n<li><strong>NCERT Solutions<\/strong>: Access detailed solutions for all chapters from the NCERT books to clear your doubts.<\/li>\n<li><strong>NCERT Exemplar Solutions<\/strong>: Get answers to tricky and high-level questions from NCERT Exemplar books.<\/li>\n<li><strong>Quick Revision Notes<\/strong>: Revise important concepts and topics quickly with easy-to-understand notes.<\/li>\n<li><strong>CBSE Guess Papers<\/strong>: Focus on probable questions with guess papers designed to give you a competitive edge.<\/li>\n<li><strong>Important Question Papers<\/strong>: Practice with question papers based on the latest CBSE syllabus and exam pattern.<\/li>\n<\/ul>\n<p><strong>Boost Your Exam Performance with myCBSEguide<\/strong><\/p>\n<p>Download <strong><a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide\">myCBSEguide<\/a> App and <a href=\"https:\/\/mycbseguide.com\/\">myCBSEGuide<\/a> Website<\/strong>\u00a0today for <strong>Class 11 sample papers<\/strong>, <strong>solutions<\/strong>, <strong>revision notes<\/strong>, and much more. Start practicing and get ready for your exams with the best study material available. For teachers, the <strong><a href=\"https:\/\/play.google.com\/store\/apps\/details?id=com.examin8.exam\">Examin8<\/a> App and <a href=\"https:\/\/examin8.com\/\">Examin8<\/a> Website<\/strong> is a powerful tool that allows you to create custom mock tests and practice papers with your own name and logo. <strong>CBSE Sample Papers Class 11 Chemistry<\/strong> are an essential tool for students looking to enhance their exam preparation and understand the exam pattern better.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>CBSE Sample Papers Class 11 Chemistry (2024-25) At myCBSEguide, we offer free downloads of CBSE Class 11 Chemistry sample papers for the 2025 academic year in PDF format. These sample papers are designed based on the latest CBSE syllabus and marking scheme for the current academic session, ensuring that students are fully prepared for their &#8230; <a title=\"CBSE Sample Papers Class 11 Chemistry 2025\" class=\"read-more\" href=\"https:\/\/mycbseguide.com\/blog\/cbse-sample-papers-class-11-chemistry\/\" aria-label=\"More on CBSE Sample Papers Class 11 Chemistry 2025\">Read more<\/a><\/p>\n","protected":false},"author":2,"featured_media":29647,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1339,1416,2032,1999],"tags":[163,1527,12,1959,1967,320,1449,1340],"class_list":["post-13676","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-cbse-sample-papers","category-chemistry","category-chemistry-sample-papers","category-class-11-sample-papers","tag-cbse-class-11","tag-cbse-question-paper","tag-cbse-sample-papers","tag-cbse-sample-papers-2024","tag-cbse-sample-papers-2025","tag-chemistry","tag-class-11","tag-model-question-papers"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.0 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>CBSE Sample Papers Class 11 Chemistry 2025 | myCBSEguide<\/title>\n<meta name=\"description\" content=\"CBSE Sample Papers Class 11 Chemistry\u00a0myCBSEguide provides 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