{"id":866005,"date":"2025-07-31T16:00:38","date_gmt":"2025-07-31T10:30:38","guid":{"rendered":"https:\/\/leverageedu.com\/discover\/?p=866005"},"modified":"2025-07-31T16:00:38","modified_gmt":"2025-07-31T10:30:38","slug":"ncert-solutions-class-11-chemistry-part-1-chapter-1-some-basic-concepts-of-chemistry","status":"publish","type":"post","link":"https:\/\/leverageedu.com\/discover\/school-education\/ncert-solutions-class-11-chemistry-part-1-chapter-1-some-basic-concepts-of-chemistry\/","title":{"rendered":"NCERT Solutions Class 11 Chemistry (Part-1) Chapter 1: Some Basic Concepts of Chemistry (Free PDF)"},"content":{"rendered":"\n<p>Class 11 Chemistry (Part-1) Chapter 1: Some Basic Concepts of Chemistry have introduced you with the importance and development of chemistry, its branches, the nature and scope of the subject, the states and classification of matter, as well as key concepts such as mass, volume, density, atomic and molecular masses, mole concept, and stoichiometry. This blog will provide you with exercises along with their solutions, which will help you understand the concepts more simply.<\/p>\n\n\n\n\n\n\n<p><strong>Explore Notes of Class 11 Chemistry<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-background has-fixed-layout\" style=\"background-color:#aadbf9\"><tbody><tr><td><strong>Chapter 1<\/strong><\/td><td><strong>Chapter 2<\/strong><\/td><td><strong>Chapter 3<\/strong><\/td><td><strong>Chapter 4<\/strong><\/td><td><strong>Chapter 5<\/strong><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-ncert-solutions-class-11-chemistry-part-1-chapter-1-some-basic-concepts-of-chemistry\">NCERT Solutions Class 11 Chemistry (Part-1) Chapter 1: Some Basic Concepts of Chemistry<\/h2>\n\n\n\n<p>Below, we have provided you with exercises mentioned in the NCERT Class 11 Chemistry (Part-1) Chapter 1: Some Basic Concepts of Chemistry, along with their solutions.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-exercises\">Exercises<\/h2>\n\n\n\n<ol class=\"wp-block-list\">\n<li>Calculate the molar mass of the following:<\/li>\n<\/ol>\n\n\n\n<p>(i) H<sub>2<\/sub>O (ii) CO<sub>2 <\/sub>(iii) CH<sub>4<\/sub><\/p>\n\n\n\n<ol start=\"2\" class=\"wp-block-list\">\n<li>Calculate the mass per cent of different elements present in sodium sulphate (Na<sub>2<\/sub>SO<sub>4<\/sub>)<\/li>\n\n\n\n<li>Determine the empirical formula of an oxide of iron, which has 69.9% iron and 30.1% dioxygen by mass.<\/li>\n\n\n\n<li>Calculate the amount of carbon dioxide that could be produced when 1.5<\/li>\n<\/ol>\n\n\n\n<p>(i) 1 mole of carbon is burnt in air.<\/p>\n\n\n\n<p>(ii) 1 mole of carbon is burnt in 16 g of dioxygen.<\/p>\n\n\n\n<p>(iii) 2 moles of carbon are burnt in 16 g of dioxygen.<\/p>\n\n\n\n<ol start=\"5\" class=\"wp-block-list\">\n<li>Calculate the mass of sodium acetate (CH<sub>3<\/sub>COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol<sup>\u20131<\/sup>.<\/li>\n\n\n\n<li>Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.41 g mL<sup>\u20131<\/sup> and the mass per cent of nitric acid in it is 69%.<\/li>\n\n\n\n<li>How much copper can be obtained from 100 g of copper sulphate (CuSO<sub>4<\/sub>)?<\/li>\n\n\n\n<li>Determine the molecular formula of an oxide of iron, in which the mass per cent of iron and oxygen are 69.9 and 30.1, respectively.<\/li>\n\n\n\n<li>Calculate the atomic mass (average) of chlorine using the following data:<\/li>\n<\/ol>\n\n\n\n<figure class=\"wp-block-table is-style-stripes\"><table class=\"has-fixed-layout\"><tbody><tr><td><\/td><td><strong>% Natural Abundance<\/strong><\/td><td><strong>Molar Mass<\/strong><\/td><\/tr><tr><td><sup>35<\/sup>Cl<\/td><td>75.77<\/td><td>34.9689<\/td><\/tr><tr><td><sup>37<\/sup>Cl<\/td><td>24.23<\/td><td>36.9659<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<ol start=\"10\" class=\"wp-block-list\">\n<li>In three moles of ethane (C<sub>2<\/sub>H<sub>6<\/sub>), calculate the following:<\/li>\n<\/ol>\n\n\n\n<p>(i) Number of moles of carbon atoms.<\/p>\n\n\n\n<p>(ii) Number of moles of hydrogen atoms.<\/p>\n\n\n\n<p>(iii) Number of molecules of ethane.<\/p>\n\n\n\n<ol start=\"11\" class=\"wp-block-list\">\n<li>What is the concentration of sugar (C<sub>12<\/sub>H<sub>22<\/sub>O<sub>11<\/sub> ) in mol L<sup>\u20131<\/sup> if 20 g are dissolved in<\/li>\n<\/ol>\n\n\n\n<p>enough water to make a final volume of up to 2 L?<\/p>\n\n\n\n<ol start=\"12\" class=\"wp-block-list\">\n<li>If the density of methanol is 0.793 kg L<sup>\u20131<\/sup>, what volume of methanol is needed to make 2.5L of its 0.25 M solution?<\/li>\n\n\n\n<li>Pressure is determined as force per unit area of the surface. The SI unit of pressure, pascal, is shown as: 1Pa = 1N m<sup>\u20132<\/sup><\/li>\n<\/ol>\n\n\n\n<p>If the mass of air at sea level is 1034 g cm<sup>\u20132<\/sup>, calculate the pressure in pascal.<\/p>\n\n\n\n<ol start=\"14\" class=\"wp-block-list\">\n<li>What is the SI unit of mass? How is it defined?<\/li>\n\n\n\n<li>What do you mean by significant figures?<\/li>\n\n\n\n<li>A sample of drinking water was found to be severely contaminated with chloroform,<\/li>\n<\/ol>\n\n\n\n<p>CHCl<sub>3<\/sub>, supposed to be carcinogenic. The level of contamination was 15 ppm (by mass).<\/p>\n\n\n\n<p>(i) Express this in per cent by mass.<\/p>\n\n\n\n<p>(ii) Determine the molality of chloroform in the water sample.<\/p>\n\n\n\n<ol start=\"17\" class=\"wp-block-list\">\n<li>Express the following in scientific notation:<\/li>\n<\/ol>\n\n\n\n<p>(i) 0.0048<\/p>\n\n\n\n<p>(ii) 234,000<\/p>\n\n\n\n<p>(iii) 8008<\/p>\n\n\n\n<p>(iv) 500.0<\/p>\n\n\n\n<p>(v) 6.0012<\/p>\n\n\n\n<ol start=\"18\" class=\"wp-block-list\">\n<li>How many significant figures are present in the following?<\/li>\n<\/ol>\n\n\n\n<p>(i) 0.0025<\/p>\n\n\n\n<p>(ii) 208<\/p>\n\n\n\n<p>(iii) 5005<\/p>\n\n\n\n<p>(iv) 126,000<\/p>\n\n\n\n<p>(v) 500.0<\/p>\n\n\n\n<p>(vi) 2.0034<\/p>\n\n\n\n<ol start=\"19\" class=\"wp-block-list\">\n<li>Round up the following up to three significant figures:<\/li>\n<\/ol>\n\n\n\n<p>(i) 34.216<\/p>\n\n\n\n<p>(ii) 10.4107<\/p>\n\n\n\n<p>(iii) 0.04597<\/p>\n\n\n\n<p>(iv) 2808<\/p>\n\n\n\n<ol start=\"20\" class=\"wp-block-list\">\n<li>The following data are obtained when dinitrogen and dioxygen react together to form different compounds:<\/li>\n<\/ol>\n\n\n\n<figure class=\"wp-block-table is-style-stripes\"><table class=\"has-fixed-layout\"><tbody><tr><td><strong>Mass of dinitrogen<\/strong><\/td><td><strong>Mass of dioxygen<\/strong><\/td><\/tr><tr><td>14 g<\/td><td>16 g<\/td><\/tr><tr><td>14 g<\/td><td>32 g<\/td><\/tr><tr><td>28 g<\/td><td>32 g<\/td><\/tr><tr><td>28 g<\/td><td>80 g<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p>Which law of chemical combination is obeyed by the above experimental data? Give<\/p>\n\n\n\n<p>Its statement.<\/p>\n\n\n\n<ol start=\"21\" class=\"wp-block-list\">\n<li>If the speed of light is 3.0 \u00d7 10 8 m s <sup>\u20131<\/sup>, calculate the distance covered by light in 2.00 ns.<\/li>\n\n\n\n<li>In a reaction<\/li>\n<\/ol>\n\n\n\n<p>A + B<sub>2<\/sub> \u2192AB<sub>2<\/sub><\/p>\n\n\n\n<p>Identify the limiting reagent, if any, in the following reaction mixtures.<\/p>\n\n\n\n<p>(i) 300 atoms of A + 200 molecules of B<\/p>\n\n\n\n<p>(ii) 2 mol A + 3 mol B<\/p>\n\n\n\n<p>(iii) 100 atoms of A + 100 molecules of B<\/p>\n\n\n\n<p>(iv) 5 mol A + 2.5 mol B<\/p>\n\n\n\n<p>(v) 2.5 mol A + 5 mol B<\/p>\n\n\n\n<ol start=\"23\" class=\"wp-block-list\">\n<li>Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation:<\/li>\n<\/ol>\n\n\n\n<p>N<sub>2<\/sub> (g) + H<sub>2<\/sub> (g)\u00a0 2NH<sub>3 <\/sub>(g)<\/p>\n\n\n\n<p>(i) Calculate the mass of ammonia produced if 2.00 \u00d7 10 3 g dinitrogen reacts with 1.00 \u00d7 10<sup>3<\/sup> g of dihydrogen.<\/p>\n\n\n\n<p>(ii) Will any of the two reactants remain unreacted?<\/p>\n\n\n\n<p>(iii) If yes, which one and what would be its mass?<\/p>\n\n\n\n<ol start=\"24\" class=\"wp-block-list\">\n<li>How are 0.50 mol Na 2CO<sub>3<\/sub> and 0.50 M Na 2CO<sub>3<\/sub> different?<\/li>\n\n\n\n<li>If 10 volumes of dihydrogen gas react with five volumes of dioxygen gas, how many volumes of water vapour would be produced?<\/li>\n\n\n\n<li>Convert the following into basic units:<\/li>\n<\/ol>\n\n\n\n<p>(i) 28.7 pm<\/p>\n\n\n\n<p>(ii) 15.15 pm<\/p>\n\n\n\n<p>(iii) 25365 mg<\/p>\n\n\n\n<ol start=\"27\" class=\"wp-block-list\">\n<li>Which one of the following will have the largest number of atoms?<\/li>\n<\/ol>\n\n\n\n<p>(i) 1 g Au (s)<\/p>\n\n\n\n<p>(ii) 1 g Na (s)<\/p>\n\n\n\n<p>(iii) 1 g Li (s)<\/p>\n\n\n\n<p>(iv) 1 g of Cl<sub>2<\/sub> (g)<\/p>\n\n\n\n<ol start=\"28\" class=\"wp-block-list\">\n<li>Calculate the molarity of a solution of ethanol in water, in which the mole fraction of Ethanol is 0.040 (assume the density of water to be one).<\/li>\n\n\n\n<li>What will be the mass of one <sup>12<\/sup>C atom in g?<\/li>\n\n\n\n<li>How many significant figures should be present in the answer to the following calculations?<\/li>\n<\/ol>\n\n\n\n<p>(i) 0.02856 \u00d7 298.15 \u00d7 0.112\/0.5785<\/p>\n\n\n\n<p>(ii) 5 \u00d7 5.364<\/p>\n\n\n\n<p>(iii) 0.0125 + 0.7864 + 0.0215<\/p>\n\n\n\n<ol start=\"31\" class=\"wp-block-list\">\n<li>Calculate the number of atoms in each of the following (i) 52 moles of Ar (ii) 52 u of He (iii) 52 g of He.<\/li>\n\n\n\n<li>A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. Calculate<\/li>\n<\/ol>\n\n\n\n<p>(i) empirical formula, (ii) molar mass of the gas, and (iii) molecular formula.<\/p>\n\n\n\n<ol start=\"33\" class=\"wp-block-list\">\n<li>Calcium carbonate reacts with aqueous HCl to give CaCl<sub>2<\/sub> and CO<sub>2<\/sub> according to the<\/li>\n<\/ol>\n\n\n\n<p>reaction, CaCO<sub>3 <\/sub>(s) + 2HCl (aq) \u2192 CaCl<sub>2<\/sub> (aq) + CO<sub>2<\/sub> (g) + H<sub>2<\/sub>O (l)<\/p>\n\n\n\n<p>What mass of CaCO<sub>3<\/sub> is required to react completely with 25 mL of 0.75 M HCl?<\/p>\n\n\n\n<ol start=\"34\" class=\"wp-block-list\">\n<li>Chlorine is prepared in the laboratory by treating manganese dioxide (MnO2) with aqueous hydrochloric acid according to the reaction<\/li>\n<\/ol>\n\n\n\n<p>4HCl (aq) + MnO<sub>2<\/sub> (s) \u2192 2H<sub>2<\/sub>O (l) + MnCl<sub>2<\/sub> (aq) + Cl<sub>2<\/sub> (g)<\/p>\n\n\n\n<p>How many grams of HCl react with 5.0 g of manganese dioxide?<\/p>\n\n\n\n<p class=\"has-pale-ocean-gradient-background has-background\"><strong>Also Read: <\/strong><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/ncert-solutions-class-11-political-science-indian-constitution-at-work-chapter-5-legislature\/\"><strong>NCERT Solutions Class 11 Political Science Indian Constitution at Work Chapter 5: Legislature (Free PDF)<\/strong><\/a><\/p>\n\n\n\n<p><strong>Solutions<\/strong><\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>Calculate the molar mass of:<\/li>\n<\/ol>\n\n\n\n<p>(i) H\u2082O<br>= 2 \u00d7 1.008 + 16.00 = 18.016 g mol\u207b\u00b9<\/p>\n\n\n\n<p>(ii) CO\u2082<br>= 12.01 + 2 \u00d7 16.00 = 44.01 g mol\u207b\u00b9<\/p>\n\n\n\n<p>(iii) CH\u2084<br>= 12.01 + 4 \u00d7 1.008 = 16.042 g mol\u207b\u00b9<\/p>\n\n\n\n<ol start=\"2\" class=\"wp-block-list\">\n<li>Mass percentage of different elements in Na\u2082SO\u2084<\/li>\n<\/ol>\n\n\n\n<p>Molar mass of Na\u2082SO\u2084 = 2 \u00d7 22.99 + 32.06 + 4 \u00d7 16.00 = 142.04 g mol\u207b\u00b9<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Na: (2 \u00d7 22.99 \/ 142.04) \u00d7 100 = 32.39%<\/li>\n\n\n\n<li>S: (32.06 \/ 142.04) \u00d7 100 = 22.57%<\/li>\n\n\n\n<li>O: (64.00 \/ 142.04) \u00d7 100 = 45.04%<\/li>\n<\/ul>\n\n\n\n<ol start=\"3\" class=\"wp-block-list\">\n<li>Empirical formula of iron oxide (69.9% Fe, 30.1% O)<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Moles of Fe = 69.9 \/ 55.85 \u2248 1.251<\/li>\n\n\n\n<li>Moles of O = 30.1 \/ 16.00 \u2248 1.881<\/li>\n<\/ul>\n\n\n\n<p>Divide by the smallest:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Fe = 1.251 \/ 1.251 = 1<\/li>\n\n\n\n<li>O = 1.881 \/ 1.251 \u2248 1.50<br>Multiply both by 2 \u2192 Fe\u2082O\u2083<\/li>\n<\/ul>\n\n\n\n<p>Empirical formula = Fe\u2082O\u2083<\/p>\n\n\n\n<ol start=\"4\" class=\"wp-block-list\">\n<li>\u00a0CO\u2082 produced by burning carbon:<\/li>\n<\/ol>\n\n\n\n<p>(i) 1 mol C + O\u2082 \u2192 CO\u2082 \u2192 1 mol CO\u2082 = 44 g<\/p>\n\n\n\n<p>(ii) 16 g O\u2082 = 0.5 mol O\u2082 \u2192 limits the reaction<\/p>\n\n\n\n<p>\u00a0C + O\u2082 \u2192 CO\u2082 \u2192 0.5 mol CO\u2082 = 0.5 \u00d7 44 = 22 g<\/p>\n\n\n\n<p>(iii) 16 g O\u2082 (0.5 mol), 2 mol C \u2192 limits reaction again \u2192 CO\u2082 = 0.5 mol = 22 g<\/p>\n\n\n\n<ol start=\"5\" class=\"wp-block-list\">\n<li>Sodium acetate for 500 mL of 0.375 M solution<\/li>\n<\/ol>\n\n\n\n<p>Moles = Molarity \u00d7 Volume (L) = 0.375 \u00d7 0.500 = 0.1875 mol<br>Mass = 0.1875 \u00d7 82.0245 = 15.38 g<\/p>\n\n\n\n<ol start=\"6\" class=\"wp-block-list\">\n<li>Given:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Density of solution = 1.41 g\/mL = 1410 g\/L<\/li>\n\n\n\n<li>Mass % of HNO\u2083 = 69%<\/li>\n\n\n\n<li>Molar mass of HNO\u2083 = 1 (H) + 14 (N) + 48 (3\u00d716 O) = 63 g\/mol<\/li>\n<\/ul>\n\n\n\n<p>Step 1: Find mass of HNO\u2083 in 1 litre of solution<br>Mass of solution = 1000 mL \u00d7 1.41 g\/mL = 1410 g<br>Mass of HNO\u2083 = 69% of 1410 =<br>= (69\/100) \u00d7 1410 = 972.9 g<\/p>\n\n\n\n<p>Step 2: Calculate the number of moles of HNO\u2083<br>Moles = Mass \/ Molar mass = 972.9 g \/ 63 g\/mol = 15.44 mol<\/p>\n\n\n\n<p>Answer: The concentration of nitric acid is 15.44 mol\/L.<\/p>\n\n\n\n<ol start=\"7\" class=\"wp-block-list\">\n<li>Given:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Mass of CuSO\u2084 = 100 g<\/li>\n\n\n\n<li>Molar mass of CuSO\u2084 =<br>Cu = 63.5, S = 32, O\u2084 = 4\u00d716 = 64 \u2192 Total = 159.5 g\/mol<\/li>\n<\/ul>\n\n\n\n<p>Step 1: Find mass % of Cu in CuSO\u2084<br>= (63.5 \/ 159.5) \u00d7 100 = 39.8%<\/p>\n\n\n\n<p>Step 2: Find the mass of copper in 100 g of CuSO\u2084<br>= (39.8 \/ 100) \u00d7 100 = 39.8 g<\/p>\n\n\n\n<p>Answer: 39.8 g of copper can be obtained from 100 g of CuSO\u2084.<\/p>\n\n\n\n<ol start=\"8\" class=\"wp-block-list\">\n<li>Given:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Fe = 69.9%, O = 30.1%<\/li>\n\n\n\n<li>Atomic mass: Fe = 55.85, O = 16<\/li>\n<\/ul>\n\n\n\n<p>Step 1: Calculate the moles of each element in 100 g<br>Fe: 69.9 \/ 55.85 = 1.251 mol<br>O: 30.1 \/ 16 = 1.881 mol<\/p>\n\n\n\n<p>Step 2: Find the simplest mole ratio<br>Divide by the smallest:<br>Fe: 1.251 \/ 1.251 = 1<br>O: 1.881 \/ 1.251 \u2248 1.50<\/p>\n\n\n\n<p>Multiply both by 2 to get whole numbers:<br>Fe : O = 2 : 3<\/p>\n\n\n\n<p>Answer: The Empirical formula is Fe\u2082O\u2083 (which is also the molecular formula).<\/p>\n\n\n\n<ol start=\"9\" class=\"wp-block-list\">\n<li>Step 1: Apply the formula for average atomic mass<br>= (75.77\/100) \u00d7 34.9689 + (24.23\/100) \u00d7 36.9659<br>= 0.7577 \u00d7 34.9689 + 0.2423 \u00d7 36.9659<br>= 26.509 + 8.957 = 35.466 u<\/li>\n<\/ol>\n\n\n\n<p>Answer: The average atomic mass of chlorine is 35.466 u.<\/p>\n\n\n\n<ol start=\"10\" class=\"wp-block-list\">\n<li>Step 1: Calculate the molar mass of sugar (C\u2081\u2082H\u2082\u2082O\u2081\u2081)<br>= (12\u00d712.01) + (22\u00d71.008) + (11\u00d716.00)<br>= 144.12 + 22.176 + 176.00 = 342.296 g\/mol<\/li>\n<\/ol>\n\n\n\n<p>Step 2: Moles of sugar =<br>= 20 g \/ 342.296 g\/mol \u2248 0.05845 mol<\/p>\n\n\n\n<p>Step 3: Molarity (mol\/L) = moles \/ volume (L)<br>= 0.05845 mol \/ 2 L = 0.0292 mol\/L<\/p>\n\n\n\n<p>Answer: 0.0292 mol\/L<\/p>\n\n\n\n<ol start=\"11\" class=\"wp-block-list\">\n<li>Step 1: Moles of CH\u2083OH needed = Molarity \u00d7 Volume =<br>0.25 mol\/L \u00d7 2.5 L = 0.625 mol<\/li>\n<\/ol>\n\n\n\n<p>Step 2: Molar mass of methanol (CH\u2083OH):<br>= 12.01 + (4\u00d71.008) + 16.00 = 32.042 g\/mol<\/p>\n\n\n\n<p>Mass required = 0.625 mol \u00d7 32.042 g\/mol =<br>= 20.03 g = 0.02003 kg<\/p>\n\n\n\n<p>Step 3: Volume = mass \/ density =<br>= 0.02003 kg \/ 0.793 kg\/L = 0.02525 L = 25.25 mL<\/p>\n\n\n\n<p>Answer: 25.25 mL<\/p>\n\n\n\n<ol start=\"12\" class=\"wp-block-list\">\n<li>If the mass of air at sea level is 1034 g cm\u207b\u00b2, calculate the pressure in pascal.<\/li>\n<\/ol>\n\n\n\n<p>Step 1: Convert mass per area to pressure<br>Force = mass \u00d7 g = 1034 g\/cm\u00b2 \u00d7 9.8 m\/s\u00b2 =<br>= 1.034 kg\/cm\u00b2 \u00d7 9.8 = 10.1332 N\/cm\u00b2<\/p>\n\n\n\n<p>Now convert cm\u00b2 to m\u00b2:<br>1 cm\u00b2 = 1\u00d710\u207b\u2074 m\u00b2 \u2192 10.1332 N\/cm\u00b2 =<br>= 10.1332 \/ (1\u00d710\u207b\u2074) = 101332 Pa<\/p>\n\n\n\n<p>Answer: Pressure = 1.013 \u00d7 10\u2075 Pa<\/p>\n\n\n\n<ol start=\"13\" class=\"wp-block-list\">\n<li>The SI unit of mass is the kilogram (kg). It is defined by taking the fixed numerical value of the Planck constant (h) as 6.62607015 \u00d7 10\u207b\u00b3\u2074 Js, which is equal to kg\u00b7m\u00b2\u00b7s\u207b\u00b9, where the metre and second are defined in terms of the speed of light and the caesium frequency.<\/li>\n\n\n\n<li>Significant figures are all the meaningful digits in a measured or calculated quantity, including all digits known with certainty and one estimated digit. They indicate the precision of a measurement.<\/li>\n\n\n\n<li>(i) 1 ppm = 1 part per million = 1 mg\/kg = 1 mg in .1,000,000 mg<br>So, 15 ppm = (15\/1,000,000) \u00d7 100 = 0.0015%<\/li>\n<\/ol>\n\n\n\n<p>Answer (i): 0.0015% by mass<\/p>\n\n\n\n<p>(ii) Assume 1 kg of water (1000 g):<\/p>\n\n\n\n<ol start=\"16\" class=\"wp-block-list\">\n<li>Mass of CHCl\u2083 = 15 mg = 0.015 g<br>Molar mass of CHCl\u2083 = 12.01 + 1.008 + (3\u00d735.45) = 119.37 g\/mol<\/li>\n<\/ol>\n\n\n\n<p>Molality = moles of solute\/mass of solvent (kg)<br>Moles of CHCl\u2083 = 0.015 \/ 119.37 \u2248 1.257 \u00d7 10\u207b\u2074 mol<br>Mass of solvent = 1 kg<\/p>\n\n\n\n<p>Answer (ii): Molality = 1.257 \u00d7 10\u207b\u2074 mol\/kg<\/p>\n\n\n\n<ol start=\"17\" class=\"wp-block-list\">\n<li>(i) 0.0048 \u2192 4.8 \u00d7 10\u207b\u00b3<br>(ii) 234,000 \u2192 2.34 \u00d7 10\u2075<br>(iii) 8008 \u2192 8.008 \u00d7 10\u00b3<br>(iv) 500.0 \u2192 5.000 \u00d7 10\u00b2<br>(v) 6.0012 \u2192 6.0012 \u00d7 10\u2070<\/li>\n\n\n\n<li>(i) 0.0025 \u2192 2<br>(ii) 208 \u2192 3<br>(iii) 5005 \u2192 4<br>(iv) 126,000 \u2192 3 (unless decimal is indicated)<br>(v) 500.0 \u2192 4<br>(vi) 2.0034 \u2192 5<\/li>\n\n\n\n<li>(i) 34.216 \u2192 34.2<br>(ii) 10.4107 \u2192 10.4<br>(iii) 0.04597 \u2192 0.0460<br>(iv) 2808 \u2192 2.81 \u00d7 10\u00b3 (or 2810 if not in scientific notation)<\/li>\n\n\n\n<li>Let\u2019s examine the ratios of masses of oxygen that combine with a fixed mass of dinitrogen.<\/li>\n<\/ol>\n\n\n\n<p>Mass of N\u2082 is fixed: 14 g<br>Ratios of oxygen masses:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Case 1: 16 g<\/li>\n\n\n\n<li>Case 2: 32 g \u2192 32\/16 = 2<\/li>\n\n\n\n<li>Case 3: N\u2082:O\u2082 = 28:32 \u2192 O\u2082 per 14 g N\u2082 = 16 g<br><\/li>\n\n\n\n<li>Case 4: 28:80 \u2192 O\u2082 per 14 g N\u2082 = 40 g<\/li>\n<\/ul>\n\n\n\n<p>So: 16 g, 32 g, 40 g \u2192 Ratio: 16:32:40 = 2:4:5<\/p>\n\n\n\n<ol start=\"21\" class=\"wp-block-list\">\n<li>Given:<br>Speed of light, c = 3.0 \u00d7 10\u2078 meters per second (m\/s)<br>Time, t = 2.00 nanoseconds (ns) = 2.00 \u00d7 10\u207b\u2079 seconds (s)<\/li>\n<\/ol>\n\n\n\n<p>Calculation:<br>Distance = Speed \u00d7 Time<br>= (3.0 \u00d7 10\u2078 m\/s) \u00d7 (2.00 \u00d7 10\u207b\u2079 s)<br>= 6.00 \u00d7 10\u207b\u00b9 meters<br>= 0.60 meters<\/p>\n\n\n\n<ol start=\"22\" class=\"wp-block-list\">\n<li>(i) 300 atoms of A + 200 molecules of B<br>Reaction: A + B\u2082 \u2192 AB\u2082 \u2192 needs 1 A per 1 B\u2082 \u2192 Limiting: A (less B\u2082 needed)<\/li>\n<\/ol>\n\n\n\n<p>(ii) 2 mol A + 3 mol B \u2192 1 mol A + 1 mol B\u2082 \u2192 2 mol A needs 2 mol B\u2082 \u2192 Only 1.5 mol B\u2082 available<br>So B is limiting<br>Limiting reagent: B<\/p>\n\n\n\n<p>(iii) 100 atoms A + 100 molecules B \u2192 100 atoms A needs 100 molecules B\u2082 \u2192Limiting: B<\/p>\n\n\n\n<p>(iv) 5 mol A + 2.5 mol B \u2192 2.5 mol B\u2082 needed \u2192 1 mol A per mol B\u2082<\/p>\n\n\n\n<p>Limiting: Neither \u2013 exact stoichiometric ratio<\/p>\n\n\n\n<p>No limiting reagent<\/p>\n\n\n\n<p>(v) 2.5 mol A + 5 mol B \u2192 5 mol B = 2.5 mol B\u2082 \u2192 1:1 with A \u2192 Both exactly match<\/p>\n\n\n\n<p>No limiting reagent<\/p>\n\n\n\n<ol start=\"23\" class=\"wp-block-list\">\n<li>Balanced chemical equation:<br>N\u2082 (g) + 3H\u2082 (g) \u2192 2NH\u2083 (g)<\/li>\n<\/ol>\n\n\n\n<p>Step 1: Calculate moles of each reactant<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Molar mass of N\u2082 = 28 g\/mol<br>Moles of N\u2082 = 2000 g \u00f7 28 g\/mol = 71.43 mol<\/li>\n\n\n\n<li>Molar mass of H\u2082 = 2 g\/mol<br>Moles of H\u2082 = 1000 g \u00f7 2 g\/mol = 500 mol<\/li>\n<\/ul>\n\n\n\n<p>Step 2: Use the mole ratio from the balanced equation<br>According to the equation:<br>1 mol N\u2082 reacts with 3 mol H\u2082 to produce 2 mol NH\u2083<\/p>\n\n\n\n<p>To react with 71.43 mol of N\u2082, we need:<br>71.43 \u00d7 3 = 214.29 mol of H\u2082<\/p>\n\n\n\n<p>But we have 500 mol of H\u2082 \u2192 excess H\u2082<br>So N\u2082 is the limiting reagent<\/p>\n\n\n\n<p>Step 3: Calculate moles of ammonia produced<br>1 mol N\u2082 produces 2 mol NH\u2083 \u2192 71.43 mol N\u2082 will produce: 71.43 \u00d7 2 = 142.86 mol NH\u2083<\/p>\n\n\n\n<p>Step 4: Calculate the mass of ammonia produced<\/p>\n\n\n\n<p>Molar mass of NH\u2083 = 17 g\/mol<br>Mass = 142.86 mol \u00d7 17 g\/mol = 2428.62 g<br>Mass of ammonia produced = 2428.62 g or 2.43 \u00d7 10\u00b3 g<\/p>\n\n\n\n<p>(ii) Will any of the two reactants remain unreacted?<\/p>\n\n\n\n<p>Yes, hydrogen will remain unreacted since it is in excess.<\/p>\n\n\n\n<p>(iii) Which one and how much of it remains unreacted?<\/p>\n\n\n\n<p>Step 1: Hydrogen required = 214.29 mol<\/p>\n\n\n\n<p>Step 2: Hydrogen available = 500 mol<\/p>\n\n\n\n<p>Excess hydrogen = 500 \u2212 214.29 = 285.71 mol<\/p>\n\n\n\n<p>Mass of leftover hydrogen = 285.71 mol \u00d7 2 g\/mol = 571.42 g<\/p>\n\n\n\n<ol start=\"24\" class=\"wp-block-list\">\n<li>0.50 mol Na\u2082CO\u2083 refers to the amount of sodium carbonate, i.e., 0.50 moles. It tells you the number of entities (molecules\/ions), not the volume or concentration. 0.50 M Na\u2082CO\u2083 refers to the concentration of a solution, i.e., 0.50 moles of sodium carbonate dissolved in 1 litre of solution.<\/li>\n\n\n\n<li>Balanced equation:<br>2H\u2082 + O\u2082 \u2192 2H\u2082O<\/li>\n<\/ol>\n\n\n\n<p>This shows that:<\/p>\n\n\n\n<p>2 volumes of hydrogen react with 1 volume of oxygen to give 2 volumes of water vapour.<\/p>\n\n\n\n<p>Now, using the same ratio:<\/p>\n\n\n\n<p>10 volumes of H\u2082 will react with 5 volumes of O\u2082 \u2192 will give 10 volumes of water vapour.<\/p>\n\n\n\n<ol start=\"26\" class=\"wp-block-list\">\n<li>(i) 28.7 pm (picometres) 1 pm = 1 \u00d7 10\u207b\u00b9\u00b2 m \u2192 28.7 pm = 2.87 \u00d7 10\u207b\u00b9\u00b9 m<\/li>\n<\/ol>\n\n\n\n<p>(ii) 15.15 pm = 1.515 \u00d7 10\u207b\u00b9\u00b9 m<\/p>\n\n\n\n<p>(iii) 25365 mg (milligrams)<br>1 mg = 1 \u00d7 10\u207b\u00b3 g = 1 \u00d7 10\u207b\u2076 kg \u2192 25365 mg = 25.365 g = 0.025365 kg<\/p>\n\n\n\n<ol start=\"27\" class=\"wp-block-list\">\n<li>Let\u2019s calculate moles of each:<\/li>\n<\/ol>\n\n\n\n<p>(i) 1 g of Au (gold)<br>Molar mass = 197 g\/mol<br>Moles = 1 \u00f7 197 = 0.00508 mol<\/p>\n\n\n\n<p>(ii) 1 g of Na (sodium)<br>Molar mass = 23 g\/mol<br>Moles = 1 \u00f7 23 = 0.0435 mol<\/p>\n\n\n\n<p>(iii) 1 g of Li (lithium)<br>Molar mass = 7 g\/mol<br>Moles = 1 \u00f7 7 = 0.143 mol<\/p>\n\n\n\n<p>(iv) 1 g of Cl\u2082 (chlorine gas)<br>Molar mass = 70.9 g\/mol<br>Moles = 1 \u00f7 70.9 = 0.0141 mol<br>Each Cl\u2082 molecule has 2 atoms \u2192 Atoms = 0.0141 \u00d7 2 = 0.0282 mol atoms<\/p>\n\n\n\n<p>Conclusion:<br>Lithium (Li) has the largest number of atoms because it has the highest number of moles per gram.<\/p>\n\n\n\n<ol start=\"28\" class=\"wp-block-list\">\n<li>Assume total moles = 1<br>Then:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Moles of ethanol = 0.040<\/li>\n\n\n\n<li>Moles of water = 1 \u2212 0.040 = 0.960<\/li>\n<\/ul>\n\n\n\n<p>Mass of water = 0.960 mol \u00d7 18 g\/mol = 17.28 g = 0.01728 kg<br>Assuming density of water = 1 g\/mL, volume = 17.28 mL = 0.01728 L<\/p>\n\n\n\n<p>Molarity = Moles of ethanol \u00f7 Volume of solution in litres<br>= 0.040 \u00f7 0.01728 = 2.31 mol\/L<\/p>\n\n\n\n<p>Final Answer:<br>Molarity = 2.31 M<\/p>\n\n\n\n<ol start=\"29\" class=\"wp-block-list\">\n<li>1 mole of \u00b9\u00b2C atoms = 12 g<br>Number of atoms in 1 mole = 6.022 \u00d7 10\u00b2\u00b3<\/li>\n<\/ol>\n\n\n\n<p>Mass of 1 atom = 12 g \u00f7 6.022 \u00d7 10\u00b2\u00b3<br>= 1.993 \u00d7 10\u207b\u00b2\u00b3 g<\/p>\n\n\n\n<ol start=\"30\" class=\"wp-block-list\">\n<li>(i) (0.02856 \u00d7 298.15 \u00d7 0.112) \u00f7 0.5785<\/li>\n<\/ol>\n\n\n\n<p>Multiply the numbers: \u2192 Result \u2248 0.165<\/p>\n\n\n\n<p>Lowest significant figures in input = 3 (0.112)<br>Answer = 0.165 (3 significant figures)<\/p>\n\n\n\n<p>(ii) 5 \u00d7 5.364 = 26.82<\/p>\n\n\n\n<p>Since \u201c5\u201d has 1 significant figure, the final answer should have 1 significant figure<\/p>\n\n\n\n<p>Answer = 30<\/p>\n\n\n\n<p>(iii) 0.0125 + 0.7864 + 0.0215 = 0.8204<\/p>\n\n\n\n<p>The number with the least decimal places is 0.0125 (4 decimals) \u2192 Keep 4 decimal places in the result<br>Answer = 0.8204<\/p>\n\n\n\n<ol start=\"31\" class=\"wp-block-list\">\n<li>52 moles of Ar<br>Atoms = 52 \u00d7 Avogadro\u2019s number = 52 \u00d7 6.022 \u00d7 10\u00b2\u00b3 = 3.13 \u00d7 10\u00b2\u2075 atoms<\/li>\n<\/ol>\n\n\n\n<p>(ii) 52 u of He<br>1 atom of He = 4 u<\/p>\n\n\n\n<p>52 u = 52 \u00f7 4 = 13 atoms<br>\u00a0<\/p>\n\n\n\n<p>(iii) 52 g of He<br>Molar mass = 4 g\/mol<br>Moles = 52 \u00f7 4 = 13 mol<br>Atoms = 13 \u00d7 6.022 \u00d7 10\u00b2\u00b3 = 7.83 \u00d7 10\u00b2\u2074 atoms<\/p>\n\n\n\n<ol start=\"32\" class=\"wp-block-list\">\n<li>Given:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>CO\u2082 formed = 3.38 g \u2192 C = (12\/44) \u00d7 3.38 = 0.922 g<\/li>\n\n\n\n<li>H\u2082O formed = 0.690 g \u2192 H = (2\/18) \u00d7 0.690 = 0.077 g<\/li>\n<\/ul>\n\n\n\n<p>Moles:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>C = 0.922 \u00f7 12 = 0.0768<\/li>\n\n\n\n<li>H = 0.077 \u00f7 1 = 0.077<\/li>\n<\/ul>\n\n\n\n<p>Ratio \u2248 C:H = 1:1 \u2192 Empirical formula = CH<\/p>\n\n\n\n<p>Given: 10 L of gas weighs 11.6 g at STP<br>Molar volume at STP = 22.4 L<br>Moles = 10 \u00f7 22.4 = 0.446 mol<br>Molar mass = 11.6 \u00f7 0.446 = ~26 g\/mol<\/p>\n\n\n\n<p>Empirical formula mass of CH = 13<br>Molecular formula = (26 \u00f7 13) = 2 \u2192 Molecular formula = C\u2082H\u2082<\/p>\n\n\n\n<ol start=\"33\" class=\"wp-block-list\">\n<li>Reaction:<br>CaCO\u2083 + 2HCl \u2192 CaCl\u2082 + CO\u2082 + H\u2082O<\/li>\n<\/ol>\n\n\n\n<p>Step 1: Moles of HCl<br>M = 0.75 mol\/L, Volume = 25 mL = 0.025 L<br>Moles of HCl = 0.75 \u00d7 0.025 = 0.01875 mol<\/p>\n\n\n\n<p>Step 2: Moles of CaCO\u2083 required<br>1 mol CaCO\u2083 reacts with 2 mol HCl \u2192 Moles of CaCO\u2083 = 0.01875 \u00f7 2 = 0.009375 mol<\/p>\n\n\n\n<p>Step 3: Mass = moles \u00d7 molar mass<br>Molar mass of CaCO\u2083 = 100 g\/mol<br>Mass = 0.009375 \u00d7 100 = 0.938 g<\/p>\n\n\n\n<ol start=\"34\" class=\"wp-block-list\">\n<li>Reaction:<br>MnO\u2082 + 4HCl \u2192 MnCl\u2082 + Cl\u2082 + 2H\u2082O<\/li>\n<\/ol>\n\n\n\n<p>Step 1: Moles of MnO\u2082<br>Molar mass = 86.9 g\/mol<br>Moles = 5.0 \u00f7 86.9 = 0.0575 mol<\/p>\n\n\n\n<p>Step 2: Moles of HCl required<br>1 mol MnO\u2082 requires 4 mol HCl \u2192 HCl = 4 \u00d7 0.0575 = 0.23 mol<\/p>\n\n\n\n<p>Step 3: Mass of HCl = moles \u00d7 molar mass = 0.23 \u00d7 36.5 = 8.40 g<\/p>\n\n\n\n<p class=\"has-pale-ocean-gradient-background has-background\"><strong>Also Read:<\/strong><a href=\"https:\/\/leverageedu.com\/discover\/school-education\/ncert-solutions-class-11-geography-fundamentals-of-physical-geography-chapter-11-world-climate-and-climate-change\/\"><strong>NCERT Solutions Class 11 Geography Fundamentals of Physical Geography Chapter 11: World Climate and Climate Change (Free PDF)<\/strong><\/a><\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-download-ncert-solutions-class-11-chemistry-part-1-chapter-1-some-basic-concepts-of-chemistry\">Download<strong> <\/strong>NCERT Solutions Class 11 Chemistry (Part-1) Chapter 1: Some Basic Concepts of Chemistry<\/h2>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-background has-fixed-layout\" style=\"background-color:#cd9ef9\"><tbody><tr><td><a href=\"https:\/\/drive.google.com\/file\/d\/13oXdKQdG6a9jfkP00rVSptXkvfjQg6lM\/view?usp=drive_link\"><strong>Download PDF of NCERT Solutions Class 11 Chemistry (Part-1) Chapter 1: Some Basic Concepts of Chemistry<\/strong><\/a><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p><strong>Download the Solutions of Other Chapters of Class 11 Psychology<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-background has-fixed-layout\" style=\"background-color:#f4e0a9\"><tbody><tr><td><strong>Chapter 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href=\"https:\/\/leverageedu.com\/discover\/school-education\/ncert-solutions-class-11-geography-fundamentals-of-physical-geography-chapter-13-movements-of-ocean-water\/\"><strong>NCERT Solutions Class 11 Geography Fundamentals of Physical Geography Chapter 13: Movements of Ocean Water (Free PDF)<\/strong><\/a><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p>For more topics, follow LeverageEdu <a href=\"https:\/\/leverageedu.com\/discover\/category\/school-education\/ncert-study-material\/\"><strong>NCERT Study Material<\/strong><\/a> today!\u00a0<\/p>\n\n\n\n<p><\/p>\n","protected":false},"excerpt":{"rendered":"Class 11 Chemistry (Part-1) Chapter 1: Some Basic Concepts of Chemistry have introduced you with the importance and&hellip;\n","protected":false},"author":133,"featured_media":866007,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"editor_notices":[],"footnotes":""},"categories":[477,389],"tags":[],"class_list":{"0":"post-866005","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-ncert-study-material","8":"category-school-education"},"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v27.5 (Yoast SEO v27.5) - https:\/\/yoast.com\/product\/yoast-seo-premium-wordpress\/ -->\n<title>NCERT Solutions Class 11 Chemistry (Part-1) Chapter 1: Some Basic Concepts of Chemistry (Free PDF) - Leverage Edu Discover<\/title>\n<meta name=\"description\" content=\"Download NCERT Solutions for Class 11 Chemistry Chapter 1: Some Basic Concepts of Chemistry (Part-1) \u2013 Free PDF with detailed step-by-step answers.\" \/>\n<meta name=\"robots\" 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