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Unit SOME BASIC CONCEP BA TS CONCEPTS 1 d OF CHEMISTRY...

Unit SOME BASIC CONCEP BA TS CONCEPTS 1 d OF CHEMISTRY CHEMISTRY he pu T I. Multiple Choice Questions (Type-I) is re ER 1. bl Two students performed the same experiment separately and each one of them recorded two readings of mass which are given below. Correct reading of mass is 3.0 g. On the basis of given data, mark the correct option out of the following statements. be C Student Readings (i) (ii) o N A 3.01 2.99 B 3.05 2.95 tt © (i) Results of both the students are neither accurate nor precise. (ii) Results of student A are both precise and accurate. (iii) Results of student B are neither precise nor accurate. (iv) Results of student B are both precise and accurate. 2. A measured temperature on Fahrenheit scale is 200 °F. What will this reading be on Celsius scale? (i) 40 °C (ii) 94 °C no (iii) 93.3 °C (iv) 30 °C 3. What will be the molarity of a solution, which contains 5.85 g of NaCl(s) per 500 mL? (i) 4 mol L –1 –1 (ii) 20 mol L  (iii) 0.2 mol L–1 (iv) 2 mol L–1 4. If 500 mL of a 5M solution is diluted to 1500 mL, what will be the molarity of the solution obtained? (i) 1.5 M (ii) 1.66 M (iii) 0.017 M (iv) 1.59 M 5. The number of atoms present in one mole of an element is equal to Avogadro d number. Which of the following element contains the greatest number of atoms? he (i) 4g He (ii) 46g Na pu T (iii) 0.40g Ca is (iv) 12g He re ER bl 6. If the concentration of glucose (C6H12O6) in blood is 0.9 g L–1, what will be the molarity of glucose in blood? (i) 5M be C (ii) 50 M (iii) 0.005 M o N (iv) 0.5 M 7. What will be the molality of the solution containing 18.25 g of HCl gas in tt © 500 g of water? (i) 0.1 m (ii) 1M (iii) 0.5 m (iv) 1m 8. One mole of any substance contains 6.022 × 1023 atoms/molecules. Number of molecules of H 2SO 4 present in 100 mL of 0.02M H2SO4 solution is ______. 20 (i) 12.044 × 10 molecules no 23 (ii) 6.022 × 10 molecules (iii) 1 × 1023 molecules 23 (iv) 12.044 × 10 molecules 9. What is the mass percent of carbon in carbon dioxide? (i) 0.034% (ii) 27.27% Exemplar Problems, Chemistry 2  (iii) 3.4% (iv) 28.7% 10. The empirical formula and molecular mass of a compound are CH2O and 180 g respectively. What will be the molecular formula of the compound? (i) C9H18O 9 (ii) CH2O (iii) C6H12O 6 (iv) C2H4O2 11. If the density of a solution is 3.12 g mL–1, the mass of 1.5 mL solution in d significant figures is _______. he (i) 4.7g (ii) 4680 × 10–3g (iii) 4.680g pu T is (iv) 46.80g re ER bl 12. Which of the following statements about a compound is incorrect? (i) A molecule of a compound has atoms of different elements. (ii) A compound cannot be separated into its constituent elements by be C physical methods of separation. (iii) A compound retains the physical properties of its constituent elements. o N (iv) The ratio of atoms of different elements in a compound is fixed. 13. Which of the following statements is correct about the reaction given below: tt © 4Fe(s) + 3O2(g) ⎯→ 2Fe2O3(g) (i) Total mass of iron and oxygen in reactants = total mass of iron and oxygen in product therefore it follows law of conservation of mass. (ii) Total mass of reactants = total mass of product; therefore, law of multiple proportions is followed. (iii) Amount of Fe2O3 can be increased by taking any one of the reactants (iron or oxygen) in excess. (iv) Amount of Fe2O3 produced will decrease if the amount of any one of the reactants (iron or oxygen) is taken in excess. no 14. Which of the following reactions is not correct according to the law of conservation of mass. (i) 2Mg(s) + O 2(g) ⎯→ 2MgO(s) (ii) C3H8(g) + O2(g) ⎯→ CO2(g) + H2O(g) (iii) P 4(s) + 5O2(g) ⎯→ P4O10(s) (iv) CH4(g) + 2O2(g) ⎯→ CO2(g) + 2H2O (g) 3 Some Basic Concepts of Chemistry  15. Which of the following statements indicates that law of multiple proportion is being followed. (i) Sample of carbon dioxide taken from any source will always have carbon and oxygen in the ratio 1:2. (ii) Carbon forms two oxides namely CO2 and CO, where masses of oxygen which combine with fixed mass of carbon are in the simple ratio 2:1. (iii) When magnesium burns in oxygen, the amount of magnesium taken for the reaction is equal to the amount of magnesium in magnesium oxide formed. (iv) At constant temperature and pressure 200 mL of hydrogen will combine with 100 mL oxygen to produce 200 mL of water vapour. d he II. Multiple Choice Questions (Type-II) In the following questions two or more options may be correct. pu T is 16. One mole of oxygen gas at STP is equal to _______. re ER (i) 6.022 × 1023 molecules of oxygen (ii) (iii) (iv) bl 6.022 × 1023 atoms of oxygen 16 g of oxygen 32 g of oxygen be C 17. Sulphuric acid reacts with sodium hydroxide as follows : o N H 2SO4 + 2NaOH ⎯→ Na2SO4 + 2H2O When 1L of 0.1M sulphuric acid solution is allowed to react with 1L of 0.1M sodium hydroxide solution, the amount of sodium sulphate formed and its tt © molarity in the solution obtained is (i) 0.1 mol L–1 (ii) 7.10 g (iii) 0.025 mol L–1 (iv) 3.55 g 18. Which of the following pairs have the same number of atoms? (i) 16 g of O2(g) and 4 g of H2(g) (ii) 16 g of O2 and 44 g of CO2 no (iii) 28 g of N2 and 32 g of O 2 (iv) 12 g of C(s) and 23 g of Na(s) 19. Which of the following solutions have the same concentration? (i) 20 g of NaOH in 200 mL of solution (ii) 0.5 mol of KCl in 200 mL of solution Exemplar Problems, Chemistry 4  (iii) 40 g of NaOH in 100 mL of solution (iv) 20 g of KOH in 200 mL of solution 20. 16 g of oxygen has same number of molecules as in (i) 16 g of CO (ii) 28 g of N2 (iii) 14 g of N2 (iv) 1.0 g of H2 21. Which of the following terms are unitless? (i) Molality d (ii) Molarity (iii) Mole fraction he (iv) Mass percent 22. One of the statements of Dalton’s atomic theory is given below: pu T is “Compounds are formed when atoms of different elements combine in a fixed ratio” re ER Which of the following laws is not related to this statement? (i) (ii) bl Law of conservation of mass Law of definite proportions be C (iii) Law of multiple proportions (iv) Avogadro law o N III. Short Answer Type tt © 23. What will be the mass of one atom of C-12 in grams? 24. How many significant figures should be present in the answer of the following calculations? 2.5 × 1.25 × 3.5 2.01 25. What is the symbol for SI unit of mole? How is the mole defined? 26. What is the difference between molality and molarity? no 27. Calculate the mass percent of calcium, phosphorus and oxygen in calcium phosphate Ca3(PO4)2. 28. 45.4 L of dinitrogen reacted with 22.7 L of dioxygen and 45.4 L of nitrous oxide was formed. The reaction is given below: 2N2(g) + O2(g) ⎯→ 2N2O(g) Which law is being obeyed in this experiment? Write the statement of the law? 5 Some Basic Concepts of Chemistry  29. If two elements can combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element, are in whole number ratio. (a) Is this statement true? (b) If yes, according to which law? (c) Give one example related to this law. 30. Calculate the average atomic mass of hydrogen using the following data : Isotope % Natural abundance Molar mass 1 H 99.985 1 2 H 0.015 2 d 31. Hydrogen gas is prepared in the laboratory by reacting dilute HCl with he granulated zinc. Following reaction takes place. Zn + 2HCl ⎯→ ZnCl 2 + H2 pu T Calculate the volume of hydrogen gas liberated at STP when 32.65 g of zinc is reacts with HCl. 1 mol of a gas occupies 22.7 L volume at STP; atomic mass of re ER Zn = 65.3 u. bl 32. The density of 3 molal solution of NaOH is 1.110 g mL–1. Calculate the molarity of the solution. be C 33. Volume of a solution changes with change in temperature, then, will the molality of the solution be affected by temperature? Give reason for your answer. o N 34. If 4 g of NaOH dissolves in 36 g of H 2O, calculate the mole fraction of each component in the solution. Also, determine the molarity of solution (specific –1 gravity of solution is 1g mL ). tt © 35. The reactant which is entirely consumed in reaction is known as limiting reagent. In the reaction 2A + 4B → 3C + 4D, when 5 moles of A react with 6 moles of B, then (i) which is the limiting reagent? (ii) calculate the amount of C formed? IV. Matching Type no 36. Match the following: (i) 88 g of CO2 (a) 0.25 mol (ii) 6.022 ×10 molecules of H2O 23 (b) 2 mol (iii) 5.6 litres of O2 at STP (c) 1 mol (iv) 96 g of O2 (d) 6.022 × 1023 molecules (v) 1 mol of any gas (e) 3 mol Exemplar Problems, Chemistry 6 37. Match the following physical quantities with units Physical quantity Unit –1 (i) Molarity (a) g mL (ii) Mole fraction (b) mol (iii) Mole (c) Pascal (iv) Molality (d) Unitless (v) Pressure (e) mol L–1 (vi) Luminous intensity (f) Candela (vii) Density (g) mol kg–1 d (viii) Mass (h) Nm–1 (i) kg he V. Assertion and Reason Type pu T is re ER In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the bl choices given below each question. 38. Assertion (A) : The empirical mass of ethene is half of its molecular mass. be C Reason (R) : The empirical formula represents the simplest whole number ratio of various atoms present in a compound. o N (i) Both A and R are true and R is the correct explanation of A. (ii) A is true but R is false. (iii) A is false but R is true. tt © (iv) Both A and R are false. 39. Assertion (A) : One atomic mass unit is defined as one twelfth of the mass of one carbon-12 atom. Reason (R) : Carbon-12 isotope is the most abundunt isotope of carbon and has been chosen as standard. (i) Both A and R are true and R is the correct explanation of A. (ii) Both A and R are true but R is not the correct explanation of A. (iii) A is true but R is false. no (iv) Both A and R are false. 40. Assertion (A) : Significant figures for 0.200 is 3 where as for 200 it is 1. Reason (R) : Zero at the end or right of a number are significant provided they are not on the right side of the decimal point. (i) Both A and R are true and R is correct explanation of A. (ii) Both A and R are true but R is not a correct explanation of A. 7 Some Basic Concepts of Chemistry  (iii) A is true but R is false. (iv) Both A and R are false. 41. Assertion (A) : Combustion of 16 g of methane gives 18 g of water. Reason (R) : In the combustion of methane, water is one of the products. (i) Both A and R are true but R is not the correct explanation of A. (ii) A is true but R is false. (iii) A is false but R is true. (iv) Both A and R are false. d VI. Long Answer Type he 42. A vessel contains 1.6 g of dioxygen at STP (273.15K, 1 atm pressure). The gas is now transferred to another vessel at constant temperature, where pu T pressure becomes half of the original pressure. Calculate is (i) volume of the new vessel. re ER (ii) number of molecules of dioxygen. bl 43. Calcium carbonate reacts with aqueous HCl to give CaCl 2 and CO 2 according to the reaction given below: CaCO3 (s) + 2HCl (aq) ⎯→ CaCl2(aq) + CO2(g) + H2O(l) be C What mass of CaCl 2 will be formed when 250 mL of 0.76 M HCl reacts with 1000 g of CaCO 3? Name the limiting reagent. Calculate the number of moles o N of CaCl 2 formed in the reaction. 44. Define the law of multiple proportions. Explain it with two examples. How tt © does this law point to the existance of atoms? 45. A box contains some identical red coloured balls, labelled as A, each weighing 2 grams. Another box contains identical blue coloured balls, labelled as B, each weighing 5 grams. Consider the combinations AB, AB2, A2B and A2B3 and show that law of multiple proportions is applicable. no Exemplar Problems, Chemistry 8 ANSWERS I. Multiple Choice Questions (Type-I) 1. (ii) 2. (iii) 3. (iii) 4. (ii) 5. (iv) 6. (iii) 7. (iv) 8. (i) 9. (ii) 10. (iii) 11. (i) 12. (iii) 13. (i) 14. (ii) 15. (ii) II. Multiple Choice Questions (Type-II) d 16. (i), (iv) 17. (ii), (iii) 18. (iii), (iv) he 19. (i), (ii) 20. (iii), (iv) 21. (iii), (iv) 22. (i), (iv) pu T is re ER III. Short Answer Type 23. 24. 2 bl 1.992648 × 10–23 g ≈ 1.99 × 10–23 g be C 25. Symbol for SI Unit of mole is mol. One mole is defined as the amount of a substance that contains as many o N particles or entities as there are atoms in exactly 12 g (0.012 kg) of the 12C isotope. tt © 26. Molality is the number of moles of solute present in one kilogram of solvent but molarity is the number of moles of solute dissolved in one litre of solution. Molality is independent of temperature whereas molarity depends on temperature. 3 × (atomic mass of calcium) 27. Mass percent of calcium = ×100 molecular mass of Ca 3 (PO4 )2 120 u = ×100 = 38.71% 310 u no 2 × (atomic mass of phosphorus) Mass percent of phosphorus = × 100 molecular mass of Ca3 (PO4 )2 2 × 31 u = ×100 = 20% 310 u 9 Some Basic Concepts of Chemistry  8 × (Atomic mass of oxygen) Mass percent of oxygen = molecular mass of Ca (PO ) × 100 3 4 2 8 × 16 u = ×100 = 41.29% 310 u 28. According to Gay Lussac’s law of gaseous volumes, gases combine or are produced in a chemical reaction in a simple ratio by volume, provided that all gases are at the same temperature and pressure. 29. (a) Yes d (b) According to the law of multiple proportions (c) H2 + O2 → H2O he 2g 16 g 18 g (c) H2 + O2 → H2O2 pu T 2g 32 g 34 g is Here masses of oxygen, (i.e., 16 g in H2O and 32 g in H2O 2) which combine re ER with fixed mass of hydrogen (2 g) are in the simple ratio i.e., 16 : 32 or 1 : 2 bl 1 {(Natural abundance of H × molar mass) + be C 2 2 (Natural abundance of H × molar mass of H)} 30. Average Atomic Mass = 100 o N 99.985 × 1 + 0.015 × 2 = 100 tt © 99.985 + 0.030 1 0 0.0 15 = = = 1.00015 u 100 100 31. From the equation, 63.5 g of zinc liberates 22.7 litre of hydrogen. So 32.65 g of zinc will liberate 22.7 L H2 22.7 32.65 g Zn × = L = 11.35 L 65.3 g Zn 2 no 32. 3 molal solution of NaOH means that 3 mols of NaOH are dissolved in 1000 g of solvent. ∴ Mass of Solution = Mass of Solvent + Mass of Solute = 1000 g + (3 × 40 g) = 1120 g 1120 Volume of Solution = mL = 1009.00 mL 1.110 Exemplar Problems, Chemistry 10  (Since density of solution = 1.110g mL–1) Since 1009 mL solution contains 3 mols of NaOH Number of moles of solute ∴ Molarity = Volume of solution in litre 3 mol = × 1000 = 2.97 M 1009.00 33. No, Molality of solution does not change with temperature since mass remains unaffected with temperature. d 34. Mass of NaOH = 4 g 4 g he Number of moles of NaOH = = 0.1 mol 40 g Mass of H2O = 36 g pu T is 36 g Number of moles of H2O = = 2 mol re ER 18 g bl Mole fraction of water = Number of moles of H2O No. of moles of water + No. of moles of NaOH be C 2 2 = = = 0.95 2 + 0.1 2.1 o N Number of moles of NaOH Mole fraction of NaOH = No. of moles of Na OH + No. of moles of water tt © 0.1 0.1 = = = 0.047 2 + 0.1 2.1 Mass of solution = mass of water + mass of NaOH = 36 g + 4 g = 40 g Volume of solution = 40 × 1 = 40 mL –1 (Since specific gravity of solution is = 1 g mL ) Number of moles of solute Molarity of solution = Volume of solution in litre no 0.1 mol NaOH = = 2.5 M 0.04 L 35. 2A + 4B → 3C + 4D According to the above equation, 2 mols of ‘A’ require 4 mols of ‘B’ for the reaction. 11 Some Basic Concepts of Chemistry  4 mol of B Hence, for 5 mols of ‘A’, the moles of ‘B’ required = 5 mol of A × 2 mol of A = 10 mol B But we have only 6 mols of ‘B’, hence, ‘B’ is the limiting reagent. So amount of ‘C’ formed is determined by amount of ‘B’. Since 4 mols of ‘B’ give 3 mols of ‘C’. Hence 6 mols of ‘B’ will give 3 mol of C 6 mol of B × = 4.5 mol of C 4 mol of B d IV. Matching Type he 36. (i) → (b) (ii) → (c) (iii) → (a) (iv) → (e) (v) → (d) pu T is 37. (i) → (e) (ii) → (d) (iii) → (b) (iv) → (g) re ER (v) → (c), (h) (vi) → (f) (vii) → (a) (viii) → (i) bl V. Assertion and Reason Type be C 38. (i) 39. (ii) 40. (iii) 41. (iii) o N VI. Long Answer Type 42. (i) p 1=1 atm, T1=273 K, V1=? tt © 32 g of oxygen occupies 22.4 L of volume at STP* 22.4 L Hence, 1.6 g of oxygen will occupy, 1.6 g oxygen × = 1.12 L 32 g oxygen V 1=1.12 L p11 p2 = = = 0.5 atm. 2 2 V 2=? no According to Boyle’s law : p 1V1 = p2V2 p1× V1 1 atm. × 1.12 L V2 = = = 2.24 L p2 0.5 atm. * Old STP conditions 273.15 K, 1 atm, volume occupied by 1 mol of gas = 22.4 L. New STP conditions 273.15 K, 1 bar, volume occupied by a gas = 22.7 L. Exemplar Problems, Chemistry 12  6.022 × 10 × 1.6 23 (ii) Number of molecules of oxygen in the vessel = 32 = 3.011 × 1022 0.76 M 43. Number of moles of HCl = 250 mL × = 0.19 mol 1000 Mass of CaCO3 = 1000 g 1000 g Number of moles of CaCO 3 = = 10 mol 100 g According to given equation 1 mol of CaCO 3 (s) requires 2 mol of HCl (aq). d Hence, for the reaction of 10 mol of CaCO3 (s) number of moles of HCl required would be: he 2 mol HCl (a q) 10 mol CaCO 3 × 1 mol CaCO 3 (s) = 20 mol HCl (aq) pu T is But we have only 0.19 mol HCl (aq), hence, HCl (aq) is limiting reagent. re ER So amount of CaCl2 formed will depend on the amount of HCl available. bl Since, 2 mol HCl (aq) forms 1 mol of CaCl2, therefore, 0.19 mol of HCl (aq) would give: 1 mol CaCl2 (aq) be C 0.19 mol HCl (aq) × = 0.095 mol 2 mol HCl (aq) or 0.095 × molar mass of CaCl2 = 0.095 × 111 = 10.54 g o N 45. (Hint : Show that the masses of B which combine with the fixed mass of A in different combinations are related to each other by simple whole numbers). tt © no 13 Some Basic Concepts of Chemistry