{"id":4129,"date":"2016-05-06T11:49:00","date_gmt":"2016-05-06T11:49:00","guid":{"rendered":"http:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-atoms-and-molecules-part-2\/"},"modified":"2018-06-07T11:34:13","modified_gmt":"2018-06-07T06:04:13","slug":"ncert-solutions-for-class-9-science-atoms-and-molecules-part-2","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-atoms-and-molecules-part-2\/","title":{"rendered":"NCERT Solutions for Class 9 Science Atoms and Molecules part 2"},"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 class=\"ez-toc-icon-toggle-span\"><svg style=\"fill: #999;color:#999\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #999;color:#999\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/span><\/span><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 eztoc-toggle-hide-by-default' ><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-atoms-and-molecules-part-2\/#NCERT_Class_9_Science_Chapter_wise_Solutions\" >NCERT Class 9 Science Chapter wise Solutions<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-atoms-and-molecules-part-2\/#NCERT_Solutions_for_Class_9_Science_Atoms_and_Molecules_part_2\" >NCERT Solutions for Class 9 Science Atoms and Molecules part 2<\/a><ul class='ez-toc-list-level-6' ><li class='ez-toc-heading-level-6'><ul class='ez-toc-list-level-6' ><li class='ez-toc-heading-level-6'><ul class='ez-toc-list-level-6' ><li class='ez-toc-heading-level-6'><ul class='ez-toc-list-level-6' ><li class='ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-atoms-and-molecules-part-2\/#1A_024_g_sample_of_compound_of_oxygen_and_boron_was_found_by_analysis_to_contain_0096_g_of_boron_and_0144_g_of_oxygen_Calculate_the_percentage_composition_of_the_compound_by_weight\" >1.A 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.096 g of boron and 0.144 g of oxygen. Calculate the percentage composition of the compound by weight.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-atoms-and-molecules-part-2\/#2When_30_g_of_carbon_is_burnt_in_800_g_oxygen_1100_g_of_carbon_dioxide_is_produced_What_mass_of_carbon_dioxide_will_be_formed_when_300_g_of_carbon_is_burnt_in_5000_g_of_oxygen_Which_law_of_chemical_combination_will_govern_your_answer\" >2.When 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen? Which law of chemical combination will govern your answer?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-atoms-and-molecules-part-2\/#3What_are_polyatomic_ions_Give_examples\" >3.What are polyatomic ions? Give examples.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-atoms-and-molecules-part-2\/#4Write_the_chemical_formulae_of_the_following\" >4.Write the chemical formulae of the following.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-7\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-atoms-and-molecules-part-2\/#5Give_the_names_of_the_elements_present_in_the_following_compounds\" >5.Give the names of the elements present in the following compounds.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-8\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-atoms-and-molecules-part-2\/#6_Calculate_the_molar_mass_of_the_following_substances\" >6. Calculate the molar mass of the following substances.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-9\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-atoms-and-molecules-part-2\/#7What_is_the_mass_of%E2%80%94\" >7.What is the mass of\u2014<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-10\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-atoms-and-molecules-part-2\/#8Convert_into_mole\" >8.Convert into mole.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-11\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-atoms-and-molecules-part-2\/#9What_is_the_mass_of\" >9.What is the mass of:<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-12\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-atoms-and-molecules-part-2\/#10Calculate_the_number_of_molecules_of_sulphur_S8_present_in_16_g_of_solid_sulphur\" >10.Calculate the number of molecules of sulphur (S8) present in 16 g of solid sulphur.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-13\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-atoms-and-molecules-part-2\/#11Calculate_the_number_of_aluminium_ions_present_in_0051_g_of_aluminium_oxide\" >11.Calculate the number of aluminium ions present in 0.051 g of aluminium oxide.<\/a><\/li><\/ul><\/li><\/ul><\/li><\/ul><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-14\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-atoms-and-molecules-part-2\/#NCERT_Solutions_for_Class_9_Science\" >NCERT Solutions for Class 9 Science<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-15\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-atoms-and-molecules-part-2\/#CBSE_app_for_Class_9\" >CBSE app for Class 9<\/a><\/li><\/ul><\/nav><\/div>\n<p>NCERT Solutions for Class 9 Science Atoms and Molecules part 2 Class 9 Science book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 9 Science chapter wise NCERT solution for Science Book for all the chapters can be downloaded from our website and myCBSEguide mobile app for free.<\/p>\n<p style=\"text-align: center;\"><strong>NCERT solutions for Science\u00a0Atoms and Molecules\u00a0<a class=\"button\" href=\"https:\/\/mycbseguide.com\/downloads\/cbse-class-09-science\/1218\/ncert-solutions\/5\/\">Download as PDF<\/a><\/strong><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/media-mycbseguide.s3.ap-south-1.amazonaws.com\/images\/blog\/09_Science_Book.jpg\" alt=\"NCERT Solutions for Class 9 Science Atoms and Molecules part 2\" width=\"154\" height=\"207\" \/><\/p>\n<h2><span class=\"ez-toc-section\" id=\"NCERT_Class_9_Science_Chapter_wise_Solutions\"><\/span><span class=\"underline-text\">NCERT Class 9 Science Chapter wise Solutions<\/span><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<ul>\n<li>01 \u2013 \u00a0Matter in Our Surroundings<\/li>\n<li>02 \u2013 Is Mattter Around us Pure<\/li>\n<li>03 \u2013 Atoms and Molecules<\/li>\n<li>04 \u2013 Structure of the Atom<\/li>\n<li>05 \u2013 The Fundamental Unit of Life<\/li>\n<li>06 \u2013 Tissues<\/li>\n<li>07 \u2013 Diversity in Living Organisms<\/li>\n<li>08 \u2013 Motion<\/li>\n<li>09 \u2013 Force and Laws of Motion<\/li>\n<li>10 \u2013 Gravitation<\/li>\n<li>11 \u2013 Word and Energy<\/li>\n<li>12 \u2013 Sound<\/li>\n<li>13 \u2013 Why Do We Fall Ill<\/li>\n<li>14 \u2013 Natural Resources<\/li>\n<li>15 \u2013 Improvement in Food Resources<\/li>\n<\/ul>\n<h2><span class=\"ez-toc-section\" id=\"NCERT_Solutions_for_Class_9_Science_Atoms_and_Molecules_part_2\"><\/span>NCERT Solutions for Class 9 Science Atoms and Molecules part 2<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"1A_024_g_sample_of_compound_of_oxygen_and_boron_was_found_by_analysis_to_contain_0096_g_of_boron_and_0144_g_of_oxygen_Calculate_the_percentage_composition_of_the_compound_by_weight\"><\/span><strong>1.A 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.096 g of boron and 0.144 g of oxygen. Calculate the percentage composition of the compound by weight.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>Mass of the given sample compound = 0.24g<\/p>\n<p style=\"text-align: justify;\">Mass of boron in the given sample compound = 0.096g<\/p>\n<p style=\"text-align: justify;\">Mass of oxygen in the given sample compound = 0.144g<\/p>\n<p style=\"text-align: justify;\">% composition of compound = % of boron and % of oxygen<\/p>\n<p style=\"text-align: justify;\">Therefore % of boron = mass of boron <img decoding=\"async\" style=\"height: 14px; width: 12px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image055.png\" \/>100\/mass of the sample compound<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 42px; width: 101px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image056.png\" \/><\/p>\n<p style=\"text-align: justify;\">= 40%<\/p>\n<p style=\"text-align: justify;\">Therefore % of oxygen = mass of oxygen <img decoding=\"async\" style=\"height: 14px; width: 12px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image055.png\" \/> 100\/mass of the sample compound<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 42px; width: 101px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image057.png\" \/><\/p>\n<p style=\"text-align: justify;\">= 60%<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 9 Science Atoms and Molecules part 2<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"2When_30_g_of_carbon_is_burnt_in_800_g_oxygen_1100_g_of_carbon_dioxide_is_produced_What_mass_of_carbon_dioxide_will_be_formed_when_300_g_of_carbon_is_burnt_in_5000_g_of_oxygen_Which_law_of_chemical_combination_will_govern_your_answer\"><\/span><strong>2.When 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen? Which law of chemical combination will govern your answer?<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>According to the law of chemical combination of constant proportions \u201cin a chemical compound the elementary constituents always combine in constant proportions by weight\/mass\u201d. Therefore, whether 3 g carbon is burnt in 8 g oxygen or 3g carbon is burnt in 50g oxygen in both cases only 11g carbon dioxide will be formed.<\/p>\n<hr \/>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"3What_are_polyatomic_ions_Give_examples\"><\/span><strong>3.What are polyatomic ions? Give examples.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>When two or more atoms in a group is having a charge, such is called a polyatomic ion. For <img decoding=\"async\" style=\"height: 25px; width: 118px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image058.png\" \/>etc.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 9 Science Atoms and Molecules part 2<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"4Write_the_chemical_formulae_of_the_following\"><\/span><strong>4.Write the chemical formulae of the following.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>(a) Magnesium chloride<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(b) Calcium oxide<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(c) Copper nitrate<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(d) Aluminium chloride<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(e) Calcium carbonate.<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" id=\"Picture 6\" style=\"height: 223px; width: 455px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image059.png\" \/><\/p>\n<hr \/>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"5Give_the_names_of_the_elements_present_in_the_following_compounds\"><\/span><strong>5.Give the names of the elements present in the following compounds.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>(a) Quick lime<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(b) Hydrogen bromide<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(c) Baking powder<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(d) Potassium sulphate.<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" id=\"Picture 7\" style=\"height: 184px; width: 560px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image060.png\" \/><\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 9 Science Atoms and Molecules part 2<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"6_Calculate_the_molar_mass_of_the_following_substances\"><\/span><strong>6. Calculate the molar mass of the following substances.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>(a) Ethyne, <img decoding=\"async\" style=\"height: 24px; width: 40px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image061.png\" \/><\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(b) Sulphur molecule, <img decoding=\"async\" style=\"height: 24px; width: 19px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image062.png\" \/><\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(c) Phosphorus molecule, <img decoding=\"async\" style=\"height: 24px; width: 19px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image063.png\" \/>(Atomic mass of phosphorus= 31)<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(d) Hydrochloric acid, HCl<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(e) Nitric acid, <img decoding=\"async\" style=\"height: 24px; width: 47px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image064.png\" \/><\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Ans. (a)<\/strong>Ethyene = <img decoding=\"async\" style=\"height: 24px; width: 39px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image065.png\" \/><img decoding=\"async\" style=\"height: 17px; width: 92px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image066.png\" \/>=24 + 2= 26 u =26 g<\/p>\n<p style=\"text-align: justify;\"><strong>(b)<\/strong>Sulphur molecular = <img decoding=\"async\" style=\"height: 24px; width: 19px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image067.png\" \/><img decoding=\"async\" style=\"height: 19px; width: 57px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image068.png\" \/>= 256 u= 256 g<\/p>\n<p style=\"text-align: justify;\"><strong>(c)<\/strong>Phosphorus molecule = <img decoding=\"async\" style=\"height: 24px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image069.png\" \/><img decoding=\"async\" style=\"height: 19px; width: 57px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image070.png\" \/> = 124 u= 124 g<\/p>\n<p style=\"text-align: justify;\"><strong>(d)<\/strong>Hydrochloric acid = HCl = 1+ 35.5 = 36.5 u= 36.5 g<\/p>\n<p style=\"text-align: justify;\"><strong>(e)<\/strong>Nitric acid = <img decoding=\"async\" style=\"height: 24px; width: 43px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image071.png\" \/>= 1 + 14 + <img decoding=\"async\" style=\"height: 22px; width: 49px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image072.png\" \/>= 15 + 48 = 63 u =63 g<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 9 Science Atoms and Molecules part 2<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"7What_is_the_mass_of%E2%80%94\"><\/span><strong>7.What is the mass of\u2014<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>(a) 1 mole of nitrogen atoms?<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(b) 4 moles of aluminium atoms (Atomic mass of aluminium= 27)?<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(c) 10 moles of sodium sulphite<img decoding=\"async\" style=\"height: 26px; width: 71px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image073.png\" \/>?<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Ans. (a)<\/strong>Atomic mass of nitrogen is 14 u.<\/p>\n<p style=\"text-align: justify;\">therefore 1 mol of N = 14g<\/p>\n<p style=\"text-align: justify;\"><strong>(b)<\/strong>Atomic mass of aluminium = 27u<\/p>\n<p style=\"text-align: justify;\">therefore 1 mol of Al = 27g and so 4 mol of Al <img decoding=\"async\" style=\"height: 22px; width: 119px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image074.png\" \/><\/p>\n<p style=\"text-align: justify;\"><strong>(c)<\/strong>molecular mass of <img decoding=\"async\" style=\"height: 24px; width: 53px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image075.png\" \/><img decoding=\"async\" style=\"height: 19px; width: 130px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image076.png\" \/>= 46 + 32 + 48 = 126 u<\/p>\n<p style=\"text-align: justify;\">therefore 1 mol of <img decoding=\"async\" style=\"height: 24px; width: 53px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image075.png\" \/>has weight\/mass 126g.<\/p>\n<p style=\"text-align: justify;\">hence, 10 mol of <img decoding=\"async\" style=\"height: 24px; width: 53px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image075.png\" \/><img decoding=\"async\" style=\"height: 19px; width: 72px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image077.png\" \/>= 1260g<\/p>\n<hr \/>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"8Convert_into_mole\"><\/span><strong>8.Convert into mole.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>(a) 12 g of oxygen gas<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(b) 20 g of water<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(c) 22 g of carbon dioxide.<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>(a)molecular mass of <img decoding=\"async\" style=\"height: 24px; width: 20px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image025.png\" \/>= 32 u= 32g(1 mole)<\/p>\n<p style=\"text-align: justify;\">since 32 g of <img decoding=\"async\" style=\"height: 24px; width: 20px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image025.png\" \/>=1mole then 12g of <img decoding=\"async\" style=\"height: 24px; width: 20px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image025.png\" \/><img decoding=\"async\" style=\"height: 42px; width: 59px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image078.png\" \/> =0.375mole.<\/p>\n<p style=\"text-align: justify;\">(b)molecular mass of <img decoding=\"async\" style=\"height: 24px; width: 34px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image011.png\" \/><img decoding=\"async\" style=\"height: 17px; width: 43px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image079.png\" \/> + 16 = 18 u= 18g(1mole)<\/p>\n<p style=\"text-align: justify;\">20g <img decoding=\"async\" style=\"height: 24px; width: 34px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image011.png\" \/><img decoding=\"async\" style=\"height: 42px; width: 59px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image080.png\" \/> = 1.11mole.<\/p>\n<p style=\"text-align: justify;\">(c)molecular mass of <img decoding=\"async\" style=\"height: 24px; width: 31px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image029.png\" \/><img decoding=\"async\" style=\"height: 19px; width: 82px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image081.png\" \/>= 12 + 32 = 44 u= 44g (1mole)<\/p>\n<p style=\"text-align: justify;\">22g of <img decoding=\"async\" style=\"height: 24px; width: 31px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image029.png\" \/><img decoding=\"async\" style=\"height: 42px; width: 59px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image082.png\" \/>= 0.5mole.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 9 Science Atoms and Molecules part 2<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"9What_is_the_mass_of\"><\/span><strong>9.What is the mass of:<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>(a) 0.2 mole of oxygen atoms?<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(b) 0.5 mole of water molecules?<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Ans. (a)<\/strong> since 1 mole of O = atomic mass of O = 16u=16g<\/p>\n<p style=\"text-align: justify;\">then 0.2mole of <img decoding=\"async\" style=\"height: 22px; width: 125px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image083.png\" \/><\/p>\n<p style=\"text-align: justify;\"><strong>(b)<\/strong>1mol of H<sub>2<\/sub>O = molecular mass of <img decoding=\"async\" style=\"height: 24px; width: 105px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image084.png\" \/>=18 u =18g<\/p>\n<p style=\"text-align: justify;\">then 0.5mol of <img decoding=\"async\" style=\"height: 24px; width: 130px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image085.png\" \/><\/p>\n<hr \/>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"10Calculate_the_number_of_molecules_of_sulphur_S8_present_in_16_g_of_solid_sulphur\"><\/span><strong>10.Calculate the number of molecules of sulphur (S8) present in 16 g of solid sulphur.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>1mol of <img decoding=\"async\" style=\"height: 24px; width: 19px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image067.png\" \/>= molecular mass of <img decoding=\"async\" style=\"height: 24px; width: 71px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image086.png\" \/>= 256u=256g<\/p>\n<p style=\"text-align: justify;\">since 256g of <img decoding=\"async\" style=\"height: 25px; width: 153px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image087.png\" \/>atoms (Avogadro number)<\/p>\n<p style=\"text-align: justify;\">16g of <img decoding=\"async\" style=\"height: 44px; width: 140px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image088.png\" \/><img decoding=\"async\" style=\"height: 22px; width: 82px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image089.png\" \/>molecules<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 9 Science Atoms and Molecules part 2<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"11Calculate_the_number_of_aluminium_ions_present_in_0051_g_of_aluminium_oxide\"><\/span><strong>11.Calculate the number of aluminium ions present in 0.051 g of aluminium oxide.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>(<em>Hint: <\/em>The mass of an ion is the same as that of an atom of the same element. Atomic mass of Al = 27 u)<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>1mol of <img decoding=\"async\" style=\"height: 24px; width: 42px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image090.png\" \/>= molecular mass of <img decoding=\"async\" style=\"height: 24px; width: 143px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image091.png\" \/>= 102u=102g<\/p>\n<p style=\"text-align: justify;\">aluminium ions present in <img decoding=\"async\" style=\"height: 25px; width: 99px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image092.png\" \/><\/p>\n<p style=\"text-align: justify;\">102 g of <img decoding=\"async\" style=\"height: 24px; width: 42px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image090.png\" \/> contains aluminium ions <img decoding=\"async\" style=\"height: 22px; width: 110px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image093.png\" \/><\/p>\n<p style=\"text-align: justify;\">then 0.051 g <img decoding=\"async\" style=\"height: 24px; width: 42px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image090.png\" \/> contains aluminium ions <img decoding=\"async\" style=\"height: 44px; width: 163px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image094.png\" \/><img decoding=\"async\" style=\"height: 22px; width: 90px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image095.png\" \/><\/p>\n<h2><span class=\"ez-toc-section\" id=\"NCERT_Solutions_for_Class_9_Science\"><\/span>NCERT Solutions for Class 9 Science<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>NCERT Solutions Class 9 Science PDF (Download) Free from myCBSEguide app and myCBSEguide website. Ncert solution class 9 Science includes text book solutions .NCERT Solutions for CBSE Class 9 Science have total 15 chapters. 9 Science NCERT Solutions in PDF for free Download on our website. Ncert Science class 9 solutions PDF and Science ncert class 9 PDF solutions with latest modifications and as per the latest CBSE syllabus are only available in myCBSEguide.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"CBSE_app_for_Class_9\"><\/span>CBSE app for Class 9<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>To download\u00a0NCERT Solutions for class 9 Science, Computer Science, Home Science,Hindi ,English, Maths Social Science do check myCBSEguide app or website. myCBSEguide provides sample papers with solution, test papers for chapter-wise practice, NCERT solutions, NCERT Exemplar solutions, quick revision notes for ready reference, CBSE guess papers and CBSE important question papers. Sample Paper all are made available through\u00a0<a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide&amp;referrer=utm_source%3Dmycbse_bottom%26utm_medium%3Dtext%26utm_campaign%3Dmycbseads\"><strong>the best app for CBSE students<\/strong><\/a>\u00a0and myCBSEguide website.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions for Class 9 Science Atoms and Molecules part 2 Class 9 Science book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text &#8230; <a title=\"NCERT Solutions for Class 9 Science Atoms and Molecules part 2\" class=\"read-more\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-atoms-and-molecules-part-2\/\" aria-label=\"More on NCERT Solutions for Class 9 Science Atoms and Molecules part 2\">Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":-1,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[281,1377],"tags":[283,1344,216,349],"class_list":["post-4129","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-ncert-solutions","category-science-cbse-class-09","tag-cbse-study-material","tag-class-9","tag-ncert-solutions","tag-science"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.0 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>NCERT Solutions for Class 9 Science Atoms and Molecules part 2<\/title>\n<meta name=\"description\" content=\"NCERT Solutions for Class 9 Science Atoms and Molecules part 2 in PDF format for free download. 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