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Std. XI Sci. Success Chemistry - I Formulae Mass of elements Atomic mass 1. One mole of atoms = 2. Mass of one atom = 3. Number of moles (n) Mass of substance = Molar mass of substance Atomic mass 6.022 × 1023 4. Number of molecules = n × Avogadro number 5. Volume of gas at STP = n × 22.414 L 6. Percentage (By weight) = Mass of the element in 1 mole of the compound × 100% Gram molecular weight of the compound 7. Molecular formula = r × empirical formula where r is ratio of molecular mass to empirical mass. Molecular mass Mass of one molecule = 6.022 × 1023 8. Solved Problems *1. Calculate the volume of hydrogen required for complete hydrogenation of 0.25 dm3 of ethyne at STP. Solution : The hydrogenation of ethyne (C2H2) is represented as, C2H2(g) + 2H2(g) → C2H6(g) ethyne ethane 1 vol 2 vol 1 vol 3 1 volume C2H6 ≡ 1 dm C2H2 requires 2 dm3H2 gas ∴ 0.25 dm3C2H2 will require = 2 × 0.25 = 0.5 dm3 H2 gas Volume of H2(g) required = 0.5 dm3. *2. 3.49 g of ammonia at STP occupies volume of 4.48 dm3 .Calculate molar mass of ammonia. Some Basic Concepts of Chemistry www.horizonpublications.in Solution : 22.4 dm3 of a gas at STP ≡ 1 mol of gas ≡ molar mass of the gas Now, 4.48 dm3 of NH3 at STP ≡ 3.49 g ∴ 22.414 dm3 of NH3 at STP 3.49 × 22.414 ≡ 4.48 = 17.46 ∴ Molar mass of NH3 = 17.46 g mol–1. *3. Calculate the volume of oxygen required for complete combustion of 0.25 mole of methane at STP. Solution : 1 mol of methane (CH4) at STP ≡ 22.414 dm3 ∴ 0.25 mol methane at STP ≡ 22.414 × 0.25 = 5.6 dm3 The combustion of CH4 is represent as, CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) 1 vol 2 vol 1 vol CH4 ≡ 1 dm3 CH4 ≡ 2 dm3O2 ∴ 5.6 dm3CH4 ≡ 2 × 5.6 = 11.2 dm3 Volume of O2 required = 11.2 dm3 at STP *4. Calculate the number of moles and molecules of ammonia present in 5.6 dm3 of its volume at STP. Solution : 22.4 dm3 of NH3 ≡ 1 mol NH3 5.6 ∴ 5.6 dm3 of NH3 ≡ = 0.25 mol NH3 22.4 1 mol NH3 contains = 6.022 × 1023 molecules of NH3 ∴ 0.25 mol NH3 contains = 6.022 × 1023 × 0.25 = 1.505 × 1023 molecules of NH3 Number of moles of NH3 = 0.25 Number of molecules of NH3 = 1.505 × 1023 *5. Calculate the mass of potassium chlorate required to liberate 6.72 dm3 of oxygen at STP, molar mass of potassium chlorate is 122.5 g/mol. Solution : Molar mass of potassium chlorate (KClO3) = 122.5 g mol–1 22.4 dm3 of O2 at STP = 1 mol O2 ∴ 6.72 dm3 of O2 at STP 1.16 HORIZON Publications Std. XI Sci. Success Chemistry - I = KClO3 → KCl + 1 mol ∴ ∴ 6.72 = 0.3 mol O2 22.4 3 O2 2 3 mol 2 3 mol O2 requires ≡ 1 mol KClO3 2 ≡ 1 mol KClO3 ≡ 122.5 g KClO3 122.5 × 0.3 0.3 mol O2 requires = 3/ 2 = 24.5 g KClO3 Mass of potassium chlorate required = 24.5 g *6. Calculate number of atoms of hydrogen present in 5.6 g of urea (molar mass of urea = 60 gmol1). Also calculate number of atoms of N, C and O. Solution : Molecular formula of mass = CO(NH2)2 = CON2H4 Molar mass of urea = 60 g mol–1 W Number of moles of urea = n = M 5.6 = = 0.09333 mol 60 Now, 1 mol urea contains 4 gram atoms of hydrogen (H) ∴ 0.09333 mol urea contains = 4 × 0.09333 = 0.3733 gram atom of H Now, 1 gm atom of H contains = 6.022 × 1023 atoms of H 0.3733 gm atom of H contains = 6.022 ×1023 × 0.3733 = 2.248 × 1023 atoms of H 1 mol of urea contains = 2 gram atoms of nitrogen (N) ∴ 0.09333 mol of urea contains = 2 × 0.09333 = 0.1866 gram atom of N ∴ Number of N atoms = 0.1866 × 6.022 × 1023 = 1.123 × 1023 N atoms 1 mol of urea contains = 1 gram atom of carbon (C) ∴ 0.09333 mol urea contains = 0.09333 gram atom of C Some Basic Concepts of Chemistry ∴ ∴ Number of C atoms = 0.09333 × 6.022 × 1023 = 0.562 × 1023 1 mol of urea contains = 1 gram atom of oxygen (O) 0.09333 mol urea contains = 0.09333 gram atom of O Number of oxygen atoms (O) = 0.09333 × 6.022 × 1023 = 0.562 × 1023 Number of H atoms = 2.248 × 1023 Number of N atoms = 1.123 × 1023 Number of C atoms = 0.562 × 1023 Number of O atoms = 0.562 × 1023 *7. Calculate the number of atoms of C, H and O in 72.5 g isopropanol, C3H7OH (molar mass 60). Solution : Molar mass of isopropanol, C3H7OH = 60 g mol–1 Mass of isopropanol = 72.5 g Number of moles of isopropanol mass of isopropanol = molar mass W ∴ n= M 72.5 = 60 = 1.208 mol 1 mol of C3H7OH ≡ 3 gram atom of C ≡ 8 gram atom of H ≡ 1 gram atom of O. ∴ Gram atom of C = 3 × 1.208 = 3.624 ∴ Number of atoms of C = 3624 × 6.022 × 1023 = 2.182 × 1024 Gram atom of H = 8 × 1.208 = 9.664 ∴ Number of atoms of H = 9.664 × 6.022 × 1023 = 5.82 × 1024 Gram atom of O = 1.208 ∴ Number of atoms of O = 1.208 × 6.022 × 1023 = 7.274 × 1023 Number of atoms of C = 2.182 × 1024 Number of atoms of H = 5.82 × 1024 Number of atoms of O = 7.274 × 1023 1.17 Std. XI Sci. Success Chemistry - I Phosphoric acid is widely used in carbonated beverages, detergents, toothpastes and fertilizers. Calculate the mass percentage of H, P and O in phosphoric acid if atomic masses are H = 1, P = 31 and O = 16. Solution : Phosphoric acid has molecular formula H3PO4. ∴ Molecular (molar) mass of H3PO4 = 3 × 1 + 31 + 4 × 16 = 98 g mol–1 Percentage composition of H in H3PO4 Atomic mass of H = × 100 Molecular mass of H 3PO 4 3 = × 100 = 3.06 % 98 Percentage composition of P in H3PO4 Atomic mass of P = × 100 Molecular mass of H 3PO 4 31 = × 100 = 31.63 % 98 ∴ Percentage of O = 100 – (3.06 + 31.63) = 65.31 % Percentage of H in H3PO4 = 3.06 % Percentage of P in H3PO4 = 31.63 % Percentage of O in H3PO4 = 65.31 % www.horizonpublications.in Percentage composition of N in HNO3 = 22.22 % Percentage composition of O in HNO3 = 76.193 % *8. *9. Calculate the mass percentage composition of the elements in nitric acid. (H = 1, N = 14, O = 16). Solution : The molecular formula of nitric acid is NHO3. Molecular mass of HNO3 = 1 + 14 + 3 × 16 = 63 g mol–1 Percentage composition of H in HNO3 Atomic mass of H = × 100 Molecular mass of HNO3 1 = × 100 = 1.587 % 63 Percentage composition of N in HNO3 Atomic mass of N = × 100 Molecular mass of HNO3 14 × 100 = 22.22% 63 Percentage composition of O in HNO3 = 100 – (1.587 + 22.22) = 76.193 % Percentage composition of H in NHO3 = 1.587 % = Some Basic Concepts of Chemistry *10. Analysis of vitamin C shows that it contains 40.92 % carbon by mass 4.58 % hydrogen and 54.50 % oxygen. Determine the empirical formula of vitamin C. Solution : Element % by mass Gram atom Atomic ratio C 40.92 40.92 12 3.41 3.41 = 3.41 =1 4.58 1 4.58 3.41 = 4.58 = 1.34 54.50 16 3.41 3.41 = 3.41 =1 H O 4.58 54.50 Whole No. ratio (Atoms) 1×3 =3 1.34 × 3 =4 1×3 =3 Hence the empirical formula of vitamin C is C3H4O3. Empirical formula of vitamin C is C3H4O3. 11. An inorganic compound contained 24.75% (w/w) potassium and 34.75% (w/w) manganese and some other common elements. Give the empirical formula of the compound. (K = 39u, Mn = 59u, O = 16u). [Intext Question text book page no. 17] Solution : Given : Atomic mass of K = 39 u, Mn = 59 u and O = 16 u Percent mass of potassium = 24.75 % Percent mass of manganese = 34.75 % Total percent mass = 59.50 % ∴ Remaining mass must be that of oxygen ∴ Percent mass of oxygen = 100 – 59.50 = 40.50% To find : The empirical formula of the given inorganic compound = ? % of potassium Moles of potassium = Atomic mass of potassium = Moles of manganese = 24.75 = 0.635 39 % of manganese Atomic mass of manganese 1.18 HORIZON Publications Std. XI Sci. Success Chemistry - I = ∴ 34.75 = 0.589 59 Moles of oxygen = % of oxygen Atomic mass of oxygen 40.50 = 2.53 16 Ratio of K : Mn : O = 0.635 0.589 2.53 : = : 1.08 : 1 : 4.29 ≈ 1 : 1 : 4 0.589 0.589 0.589 The empirical formula of given inorganic compound is KMnO4. = ∴ Phosphoric acid used in carbonated beverages contain 3.086% (w/w) hydrogen, 31.61% (w/w) phosphorous and remaining oxygen. If atomic masses of hydrogen, phosphorous and oxygen are 1.01 u, 31.o u and 16 u respectively and if molar mass of phosphoric acid is 98.03 g mol-1, what is the molecular formula of phosphoric acid ? [Intext Question text book page no. 17] Solution : Given : Atomic mass of H = 1.01 u, P = 31.0 u and O = 16 u Percent mass of hydrogen = 3.086 % Percent mass of phosphorous = 31.61 % ∴ Total percent mass = 34.696 % ∴ Remaining amount is oxygen ∴ Percent mass of oxygen = 100 – 34.696 = 65.304 % To find : The molecular formula of phosphoric acid = ? Moles of hydrogen = % mass of hydrogen ∴ 12. Multiple Choice Questions 1. 2. 3. 4. Atomic mass of hydrogen = 3.086 = 3.05 1.01 Moles of phosphorous = = % mass of phosphorous Atomic mass of phosphorous 31.61 = 1.019 31 Moles of Oxygen = % mass of oxygen Atomic mass of oxygen 65.304 = 4.08 16 Ratio of H : P : O 3.05 1.019 4.08 = : : = 3 :1: 4 1.019 1.019 1.019 Hence, the empirical formula of phosphoric acid = H3PO4 = ∴ Some Basic Concepts of Chemistry Empirical formula mass = 3 (1.01) + 1 (31) + 4 (16) = 3.03 + 31 + 64 = 98.03 g mol–1 Molar mass 98.03 r= = = 1 Empirical formula mass 98.03 Molecular formula = r × Empirical formula = l × H3PO4 = H3PO4 The molecular formula of phosphoric acid is H3PO4. 5. 6. 7. 8. The gas law was formulated by (a) Avogadro (b) Gay-Lussac (c) Aristotle (d) Joseph Priestley The branch of chemistry which deals with carbon compound is called …… chemistry. (a) organic (b) inorganic (c) carbon (d) bio …… can not be carried out in a lab. (a) Photosynthesis (b) Reduction (c) Oxidation (d) Hydration How may moles of electron weigh one kilogram? (a) 6.022 × 1022 1 (b) ×1031 9.108 6.022 (c) ×1054 9.108 1 (d) ×108 9.108 × 6.022 What is the mass of 0.5 mole of ozone molecules ? (a) 8 g (b) 16 g (c) 24 g (d) 48 g Percentage of nitrogen in urea is about (a) 46 % (b) 85 % (c) 18 % (d) 28 % Which of the following has same moleculear formula and empirical formula ? (b) C6H12O6 (a) CO2 (c) C2H4 (d) all of these The empirical formula of C2H2 is……. (a) C2H4 (b) CH (c) CH4 (d) all of these 1.19 Std. XI Sci. Success Chemistry - I 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. …… is the quantitative relationship between the reactants and products in a balanced chemical equation. (a) Stiochiometry (b) Complexometry (c) Chemistry (d) Reactions If 0.5 mol of BaCl2 is mixed with 0.2 mole of Na3PO4, the maximum number of moles of Ba3(PO4)2 that can be formed is (a) 0.7 (b) 0.5 (c) 0.3 (d) 0.1 When oxygen forms ozone, the atomicity changes from (a) 1 to 2 (b) 1 to 3 (c) 2 to 3 (d) 2 to 4 3 If 0.0224 dm of a gas at STP weigths 0.06g , the moleuclar mass of the gas will be (a) 6 (b) 600 (c) 30 (d) 60 6.4 mg of sulphur on comustion will give number of molecules of SO2 equal to (a) 1.2044 × 1020 (b) 6.4 × 1023 (c) 0.2 (d) 3.2 × 1021 0.01 mol of NH4NO3 will weigh (Atomic mass, H = 1m N = 14, O = 16) (a) 6.4 × 10−3 kg (b) 3.4 × 10−4 kg (c) 7.2 × 10−3 kg (d) 8 × 10−4 kg The number of atoms of He present in 104 u of He is (a) 13 (b) 26 –3 (c) 3.5 × 10 kg (d) 8 × 10–4 kg Mole is the SI unit of ……… (a) Volume (b) Pressure (c) Amount of substance (d) Density Mixture of solids constitute …… system. (a) Homogeneous (b) Heterogenous (c) Either (a) or (b) (d) Neither (a) not (b) Two elements, A and B, combine to form a compound in which ‘a’ g of A combines with ‘b1’ and ‘b2’ g of B respectively. According to law of multiple proportion …….. (a) b1 = b2 (b) b1 and b2 bear a simple whole number ratio (c) a1 and b1 bear a whole number ratio (d) no relation exists between b1 and b2 Some Basic Concepts of Chemistry www.horizonpublications.in 19. 20. 21. 22. 23. Molecule is the smallest particle of ….. (a) compound (b) substance (c) mixture (d) element One mole of oxygen weighs ……… (a) 8 g (b) 32 g (c) 16 g (d) 6.023 × 1023 g The number of atoms present in a molecule of a substance is called ………. (a) Atomicity (b) Volume (c) Density (d) Mass The number of molecules present in 8 g of oxygen gas are ……… (b) 3.011 × 1023 (a) 6.022 × 1023 23 (c) 12.044 × 10 (d) 1.505 × 1023 Atomicity of mercury vapour is …….. (a) 1 (b) 2 (c) 3 (d) 4 1. 6. 11. 16. 21. b a c c a 2. 7. 12. 17. 22. a a d b d Answers 3. a 8. b 13. a 18. b 23. a 4. 9. 14. 19. d a d a 5. 10. 15. 20. c d b b 1.20