{"id":4128,"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-1\/"},"modified":"2018-06-07T17:15:33","modified_gmt":"2018-06-07T11:45:33","slug":"ncert-solutions-for-class-9-science-atoms-and-molecules-part-1","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-atoms-and-molecules-part-1\/","title":{"rendered":"NCERT Solutions for Class 9 Science Atoms and Molecules part 1"},"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-1\/#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-1\/#NCERT_Solutions_for_Class_9_Science_Atoms_and_Molecules_part_1\" >NCERT Solutions for Class 9 Science Atoms and Molecules part 1<\/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-1\/#1_In_a_reaction_53_g_of_sodium_carbonate_reacted_with_6_g_of_ethanoic_acid_The_products_were_22_g_of_carbon_dioxide_09_g_water_and_82_g_of_sodium_ethanoate_Show_that_these_observations_are_in_agreement_with_the_law_of_conservation_of_mass\" >1. In a reaction, 5.3 g of sodium carbonate reacted with 6 g of ethanoic acid. The products were 2.2 g of carbon dioxide, 0.9 g water and 8.2 g of sodium ethanoate. Show that these observations are in agreement with the law of conservation of mass.<\/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-1\/#2Hydrogen_and_oxygen_combine_in_the_ratio_of_1_8_by_mass_to_form_water_What_mass_of_oxygen_gas_would_be_required_to_react_completely_with_3_g_of_hydrogen_gas\" >2.Hydrogen and oxygen combine in the ratio of 1:8 by mass to form water. What mass of oxygen gas would be required to react completely with 3 g of hydrogen gas?<\/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-1\/#3Which_postulate_of_Daltons_atomic_theory_is_the_result_of_the_law_of_conservation_of_mass\" >3.Which postulate of Dalton\u2019s atomic theory is the result of the law of conservation of mass?<\/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-1\/#4Which_postulate_of_Daltons_atomic_theory_can_explain_the_law_of_definite_proportions\" >4.Which postulate of Dalton\u2019s atomic theory can explain the law of definite proportions?<\/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-1\/#1Define_the_atomic_mass_unit\" >1.Define the atomic mass unit.<\/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-1\/#2Why_is_it_not_possible_to_see_an_atom_with_naked_eyes\" >2.Why is it not possible to see an atom with naked eyes?<\/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-1\/#2Write_down_the_names_of_compounds_represented_by_following_formulae\" >2.Write down the names of compounds represented by following formulae:<\/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-1\/#3What_is_meant_by_the_term_chemical_formula\" >3.What is meant by the term chemical formula?<\/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-1\/#4How_many_atoms_are_present_in_a\" >4.How many atoms are present in a<\/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-1\/#1_Calculate_the_molecular_masses_of\" >1. Calculate the molecular masses of<\/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-1\/#2_Calculate_the_formula_unit_masses_of_given_atomic_masses_of_Zn_65_u_Na_23_u_K_39_u_C_12_u_and_O_16_u\" >2. Calculate the formula unit masses of, given atomic masses of Zn = 65 u, Na = 23 u, K = 39 u, C = 12 u, and O = 16 u.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-14\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-atoms-and-molecules-part-1\/#1_If_one_mole_of_carbon_atoms_weighs_12_grams_what_is_the_mass_in_grams_of_1_atom_of_carbon\" >1. If one mole of carbon atoms weighs 12 grams, what is the mass (in grams) of 1 atom of carbon?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-15\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-atoms-and-molecules-part-1\/#2Which_has_more_number_of_atoms_100_grams_of_sodium_or_100_grams_of_iron_given_atomic_mass_of_Na_23_u_Fe_56_u\" >2.Which has more number of atoms, 100 grams of sodium or 100 grams of iron (given, atomic mass of Na = 23 u, Fe = 56 u)?<\/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-16\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-atoms-and-molecules-part-1\/#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-17\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-atoms-and-molecules-part-1\/#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 1 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 1\" width=\"171\" height=\"230\" \/><\/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 Matter 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_1\"><\/span>NCERT Solutions for Class 9 Science Atoms and Molecules part 1<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"1_In_a_reaction_53_g_of_sodium_carbonate_reacted_with_6_g_of_ethanoic_acid_The_products_were_22_g_of_carbon_dioxide_09_g_water_and_82_g_of_sodium_ethanoate_Show_that_these_observations_are_in_agreement_with_the_law_of_conservation_of_mass\"><\/span><strong>1. In a reaction, 5.3 g of sodium carbonate reacted with 6 g of ethanoic acid. The products were 2.2 g of carbon dioxide, 0.9 g water and 8.2 g of sodium ethanoate. Show that these observations are in agreement with the law of conservation of mass.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>sodium carbonate + ethanoic acid \u2192 sodium ethanoate + carbon dioxide + water<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>According to law of conservation of mass:<\/p>\n<p style=\"text-align: justify;\">mass of reactants = mass of products<\/p>\n<p style=\"text-align: justify;\">Lets calculate and find out both results \u2013<\/p>\n<p style=\"text-align: justify;\">mass of reactants = mass of sodium carbonate +mass of ethanoic acid<\/p>\n<p style=\"text-align: justify;\">= 5.3g + 6g<\/p>\n<p style=\"text-align: justify;\">= 11.3g<\/p>\n<p style=\"text-align: justify;\">mass of products = mass of sodium ethanoate + mass of carbon dioxide + mass of water<\/p>\n<p style=\"text-align: justify;\">= 8.2g +2.2g + 0.9g = 11.3g<\/p>\n<p style=\"text-align: justify;\">Hence it is proved that these observations are in agreement with the law of conservation of mass.<\/p>\n<hr \/>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"2Hydrogen_and_oxygen_combine_in_the_ratio_of_1_8_by_mass_to_form_water_What_mass_of_oxygen_gas_would_be_required_to_react_completely_with_3_g_of_hydrogen_gas\"><\/span><strong>2.Hydrogen and oxygen combine in the ratio of 1:8 by mass to form water. What mass of oxygen gas would be required to react completely with 3 g of hydrogen gas?<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>As per the given 1:8 ratio mass of oxygen gas required to react completely with 1g of hydrogen gas is 8g.<\/p>\n<p style=\"text-align: justify;\">Therefore, mass of oxygen gas required to react completely with 3g of hydrogen gas will be = <img decoding=\"async\" style=\"height: 22px; width: 79px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image001.png\" \/><\/p>\n<hr \/>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"3Which_postulate_of_Daltons_atomic_theory_is_the_result_of_the_law_of_conservation_of_mass\"><\/span><strong>3.Which postulate of Dalton\u2019s atomic theory is the result of the law of conservation of mass?<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>The postulate of Dalton\u2019s atomic theory which is the result of the law of conservation of mass is mentioned as below:<\/p>\n<p style=\"text-align: justify;\">Atoms are indivisible particles, which cannot be created or destroyed in a chemical reaction.<\/p>\n<hr \/>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"4Which_postulate_of_Daltons_atomic_theory_can_explain_the_law_of_definite_proportions\"><\/span><strong>4.Which postulate of Dalton\u2019s atomic theory can explain the law of definite proportions?<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>The postulate of Dalton\u2019s atomic theory which explains the law of definite proportions is \u201cAtoms combine in the ratio of small whole numbers to form compounds and the relative number and kinds of atoms are constant in a given compound.\u201d<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 9 Science Atoms and Molecules part 1<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"1Define_the_atomic_mass_unit\"><\/span><strong>1.Define the atomic mass unit.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>According to the latest recommendations of International Union of Pure and Applied Chemistry (IUPAC) the atomic mass unit (amu) is abbreviated as u or unified mass.<\/p>\n<p style=\"text-align: justify;\">For chemical calculations the atomic masses of elements are expressed by taking the atomic mass of one atom of an element as the standard mass. Like the atomic mass of carbon is taken as 12 units and each unit is called as 1 a.m.u i.e.<\/p>\n<p style=\"text-align: justify;\">1 amu = 1\/12 of atomic masses of <img decoding=\"async\" style=\"height: 25px; width: 32px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image002.png\" \/>.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 9 Science Atoms and Molecules part 1<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"2Why_is_it_not_possible_to_see_an_atom_with_naked_eyes\"><\/span><strong>2.Why is it not possible to see an atom with naked eyes?<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans.<\/strong> An atom is an extremely minute particle and as such actual mass of an atom of hydrogen is considered to be<img decoding=\"async\" style=\"height: 24px; width: 77px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image003.png\" \/>. That is why it is not possible to see an atom with naked eyes.<\/p>\n<hr \/>\n<p style=\"text-align: justify;\"><strong>(Page No. 39)<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>1.Write down the formulae of<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(i)sodium oxide<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(ii)aluminium chloride<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(iii)sodium Sulphide<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(iv)magnesium hydroxide<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" id=\"Picture 1\" style=\"height: 156px; width: 352px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image004.png\" \/><\/p>\n<hr \/>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"2Write_down_the_names_of_compounds_represented_by_following_formulae\"><\/span><strong>2.Write down the names of compounds represented by following formulae:<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>(i)<img decoding=\"async\" style=\"height: 26px; width: 77px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image005.png\" \/><\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(ii)<img decoding=\"async\" style=\"height: 24px; width: 45px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image006.png\" \/><\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(iii)<img decoding=\"async\" style=\"height: 24px; width: 51px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image007.png\" \/><\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(iv)<img decoding=\"async\" style=\"height: 24px; width: 47px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image008.png\" \/><\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(v)<img decoding=\"async\" style=\"height: 24px; width: 53px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image009.png\" \/>.<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" id=\"Picture 5\" style=\"height: 183px; width: 377px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image010.png\" \/><\/p>\n<hr \/>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"3What_is_meant_by_the_term_chemical_formula\"><\/span><strong>3.What is meant by the term chemical formula?<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>A chemical formula is the representation of elements present in a compound with the help of symbols and also the number of atoms of each element with those numbers only. for eg: A molecule of water (compound) contains 2 atoms of hydrogen and one atom of oxygen hence its chemical formula is<img decoding=\"async\" style=\"height: 24px; width: 34px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image011.png\" \/>.<\/p>\n<hr \/>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"4How_many_atoms_are_present_in_a\"><\/span><strong>4.How many atoms are present in a<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>(i)<img decoding=\"async\" style=\"height: 24px; width: 32px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image012.png\" \/>molecule and<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(ii)<img decoding=\"async\" style=\"height: 25px; width: 44px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image013.png\" \/>ion?<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Ans.<\/strong> <strong>(i)<\/strong> 2 atoms of hydrogen + 1 atom of sulphur = 3 atoms<\/p>\n<p style=\"text-align: justify;\"><strong>(ii)<\/strong>1 atom of phosphorus + 4 atoms of oxygen = 5 atoms<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 9 Science Atoms and Molecules part 1<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"1_Calculate_the_molecular_masses_of\"><\/span><strong>1. Calculate the molecular masses of <\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong><img decoding=\"async\" style=\"height: 24px; width: 23px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image014.png\" \/><\/strong><\/p>\n<p style=\"text-align: justify;\"><strong><img decoding=\"async\" style=\"height: 24px; width: 23px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image015.png\" \/><\/strong><\/p>\n<p style=\"text-align: justify;\"><strong><img decoding=\"async\" style=\"height: 24px; width: 26px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image016.png\" \/><\/strong><\/p>\n<p style=\"text-align: justify;\"><strong><img decoding=\"async\" style=\"height: 24px; width: 34px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image017.png\" \/><\/strong><\/p>\n<p style=\"text-align: justify;\"><strong><img decoding=\"async\" style=\"height: 24px; width: 34px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image018.png\" \/><\/strong><\/p>\n<p style=\"text-align: justify;\"><strong><img decoding=\"async\" style=\"height: 24px; width: 40px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image019.png\" \/><\/strong><\/p>\n<p style=\"text-align: justify;\"><strong><img decoding=\"async\" style=\"height: 24px; width: 40px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image020.png\" \/><\/strong><\/p>\n<p style=\"text-align: justify;\"><strong><img decoding=\"async\" style=\"height: 24px; width: 34px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image021.png\" \/><\/strong><\/p>\n<p style=\"text-align: justify;\"><strong><img decoding=\"async\" style=\"height: 24px; width: 60px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image022.png\" \/><\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>Molecular mass of <img decoding=\"async\" style=\"height: 24px; width: 23px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image023.png\" \/>= atomic mass of<img decoding=\"async\" style=\"height: 19px; width: 113px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image024.png\" \/>.<\/p>\n<p style=\"text-align: justify;\">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\" \/>= atomic mass of<img decoding=\"async\" style=\"height: 19px; width: 128px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image026.png\" \/>.<\/p>\n<p style=\"text-align: justify;\">Molecular mass of <img decoding=\"async\" style=\"height: 24px; width: 24px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image027.png\" \/>= atomic mass of<img decoding=\"async\" style=\"height: 19px; width: 144px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image028.png\" \/>.<\/p>\n<p style=\"text-align: justify;\">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\" \/>= atomic mass of C + (atomic mass of O x 2)<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 26px; width: 110px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image030.png\" \/><\/p>\n<p style=\"text-align: justify;\">= (12 + 32)<\/p>\n<p style=\"text-align: justify;\">= 44 u<\/p>\n<p style=\"text-align: justify;\">Molecular mass of <img decoding=\"async\" style=\"height: 24px; width: 34px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image031.png\" \/>= 12 + atomic mass of hydrogen <img decoding=\"async\" style=\"height: 17px; width: 22px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image032.png\" \/><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 22px; width: 88px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image033.png\" \/><\/p>\n<p style=\"text-align: justify;\">= 12 + 4<\/p>\n<p style=\"text-align: justify;\">= 16 u<\/p>\n<p style=\"text-align: justify;\">Molecular mass of <img decoding=\"async\" style=\"height: 24px; width: 39px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image034.png\" \/><img decoding=\"async\" style=\"height: 22px; width: 113px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image035.png\" \/>= 24+6 = 30 u<\/p>\n<p style=\"text-align: justify;\">Molecular mass of <img decoding=\"async\" style=\"height: 24px; width: 39px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image036.png\" \/><img decoding=\"async\" style=\"height: 22px; width: 113px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image037.png\" \/>= 24 + 4 = 28 u<\/p>\n<p style=\"text-align: justify;\">Molecular mass of<img decoding=\"async\" style=\"height: 24px; width: 34px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image038.png\" \/> <img decoding=\"async\" style=\"height: 22px; width: 82px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image039.png\" \/>= 14 + 3= 17 u<\/p>\n<p style=\"text-align: justify;\">Molecular mass of <img decoding=\"async\" style=\"height: 24px; width: 57px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image040.png\" \/><img decoding=\"async\" style=\"height: 22px; width: 130px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image041.png\" \/>= 12+3+16+1 = 32 u<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 9 Science Atoms and Molecules part 1<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"2_Calculate_the_formula_unit_masses_of_given_atomic_masses_of_Zn_65_u_Na_23_u_K_39_u_C_12_u_and_O_16_u\"><\/span><strong>2. Calculate the formula unit masses of<img decoding=\"async\" style=\"height: 24px; width: 144px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image042.png\" \/>, given atomic masses of Zn = 65 u, Na = 23 u, K = 39 u, C = 12 u, and O = 16 u.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>Formula unit mass of:<\/p>\n<p style=\"text-align: justify;\"><strong>i)<\/strong> ZnO = Atomic mass of Zn + atomic mass of O = (65 + 16) u = 81 u<\/p>\n<p style=\"text-align: justify;\"><strong>ii)<\/strong> <img decoding=\"async\" style=\"height: 24px; width: 42px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image043.png\" \/>= Atomic mass of Na + atomic mass of <img decoding=\"async\" style=\"height: 22px; width: 108px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image044.png\" \/>= 46 + 16 = 62 u<\/p>\n<p style=\"text-align: justify;\"><strong>iii) <img decoding=\"async\" style=\"height: 24px; width: 48px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image045.png\" \/><img decoding=\"async\" style=\"height: 22px; width: 152px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image046.png\" \/><\/strong> = 78 + 12 + 48 = 138 u<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 9 Science Atoms and Molecules part 1<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"1_If_one_mole_of_carbon_atoms_weighs_12_grams_what_is_the_mass_in_grams_of_1_atom_of_carbon\"><\/span><strong>1. If one mole of carbon atoms weighs 12 grams, what is the mass (in grams) of 1 atom of carbon?<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>Weight of one mole of carbon = atomic mass of carbon (1 atom of carbon) = 12 u<\/p>\n<p style=\"text-align: justify;\">Therefore, one mole of carbon contains = 12 g <img decoding=\"async\" style=\"height: 22px; width: 95px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image047.png\" \/>atoms (Avogadro number)<\/p>\n<p style=\"text-align: justify;\">so 1 atom of carbon = 12\/ g<\/p>\n<p style=\"text-align: justify;\">or, 12 <img decoding=\"async\" style=\"height: 42px; width: 116px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image048.png\" \/><\/p>\n<p style=\"text-align: justify;\">1 <img decoding=\"async\" style=\"height: 42px; width: 145px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image049.png\" \/><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 42px; width: 127px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image050.png\" \/><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 24px; width: 161px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image051.png\" \/><\/p>\n<p style=\"text-align: justify;\">or, <img decoding=\"async\" style=\"height: 24px; width: 153px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image052.png\" \/><\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 9 Science Atoms and Molecules part 1<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"2Which_has_more_number_of_atoms_100_grams_of_sodium_or_100_grams_of_iron_given_atomic_mass_of_Na_23_u_Fe_56_u\"><\/span><strong>2.Which has more number of atoms, 100 grams of sodium or 100 grams of iron (given, atomic mass of Na = 23 u, Fe = 56 u)?<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>We can find out the element with more number of atoms by calculating number of moles of each of them:<\/p>\n<p style=\"text-align: justify;\">Number of moles of sodium in <img decoding=\"async\" style=\"height: 24px; width: 100px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image053.png\" \/>= 100\/23 = 4.34<\/p>\n<p style=\"text-align: justify;\">Number of moles of iron in <img decoding=\"async\" style=\"height: 24px; width: 104px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/09\/science\/ch03\/image054.png\" \/>= 100\/56 = 1.79<\/p>\n<p style=\"text-align: justify;\">Therefore, the number of atoms is more for sodium as compared to iron.<\/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 1 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 1\" class=\"read-more\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-9-science-atoms-and-molecules-part-1\/\" aria-label=\"More on NCERT Solutions for Class 9 Science Atoms and Molecules part 1\">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-4128","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 1<\/title>\n<meta name=\"description\" content=\"NCERT Solutions for Class 9 Science Atoms and Molecules part 1 in PDF format for free download. 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