1. Which one of the following pairs of compounds illustrate the law of multiple proportions?
(a) H2 S and SO2 (b) NH3 and NCl3 (c) FeCl2 and FeCl3 (d) CuO and Cu2 O (d ) The law of multiple proportions is best illustrated by CuO and Cu2 O where copper combines with oxygen in different ratios (1:1 and 2:1). 2. The hydrogen phosphate of certain metal has formula MHPO4 . The formula of metal chloride would be
(a) MCl (b) M2 Cl2 (c) MCl2 (d) MCl3 (c ) Since the phosphate is MHPO4 , the metal has +2 valency (as HPO4 2- ), so chloride will be MCl2 . 3. Modern atomic weight scale is based on
(a) H-1 (b) C-12 (c) O-16 (d) C-14 (b ) The modern atomic weight scale uses C-12 as the standard reference (exactly 12 amu). 4. 1 amu is equal to
(a) 1/12 of C-12 (b) 1 g of H (c) 1/14 of O-16 (d) 1.66 × 10-23 kg (d ) 1 amu = 1.66 × 10-27 kg = 1.66 × 10-24 g. 5. The ratio of the rates of diffusion of a given element to that of helium is 4. What will be the molecular weight of the element?
(a ) Using Graham's law: r1 /r2 = √(M2 /M1 ). Here, 4 = √(4/M) → M = 0.25. 6. The law of reciprocal proportions can be used to determine
(a) atomic weight of a gas (b) equivalent weights (c) molecular weights of gases (d) none of the above (b ) The law of reciprocal proportions helps determine equivalent weights of elements. 7. The specific heat of a metal is 0.16, its approximate atomic weight would be
(b ) Using Dulong-Petit's law: Approx. atomic weight = 6.4/specific heat = 6.4/0.16 = 40. 8. If the equivalent weight of a trivalent metal is 32.7, the molecular weight of its chloride is
(a) 68.2 (b) 103.7 (c) 204.6 (d) 32.7 (c ) Atomic weight = Eq. wt × valency = 32.7 × 3 = 98.1. Chloride formula = MCl3 → Mol. wt. = 98.1 + (3 × 35.5) = 204.6. 9. Vapour density of a gas is 22. Its molecular weight will be
(c ) Molecular weight = 2 × Vapour density = 2 × 22 = 44. 10. In the reaction: 2Na2 S2 O3 + I2 → Na2 S4 O6 + 2NaI, the equivalent weight of Na2 S2 O3 (mol. wt. = M) is equal to
(a ) In this reaction, Na2 S2 O3 changes oxidation state by 1 unit per molecule, so equivalent weight = M/1 = M. 11. A metallic oxide contains 60% of the metal. The equivalent weight of the metal is
(b ) Let metal weight = 60g, oxygen = 40g. Eq. wt. of metal = (60 × 8)/40 = 12. But since oxide is MO (as 60:40 ≈ 3:2 ratio suggests M2 O3 ), valency is 3 → Atomic wt. = 12 × 3 = 36. (Note: Question seems inconsistent with options). 12. The equivalent weight of KMnO4 (in acid medium) is (At. wt. of K = 39, Mn = 55)
(a) 158 (b) 15.8 (c) 31.6 (d) 3.16 (c ) In acid medium, MnO4 - → Mn2+ (change of 5 electrons). Mol. wt. = 158, so Eq. wt. = 158/5 = 31.6. 13. M is the molecular weight of KMnO4 . The equivalent weight of KMnO4 when it is converted into K2 MnO4 is
(a) M (b) M/3 (c) M/5 (d) M/7 (a ) In conversion to K2 MnO4 , Mn changes from +7 to +6 (1 electron change), so Eq. wt. = M/1 = M. 14. In the reaction: SO2 + 2H2 S → 3S + 2H2 O, which choice has value twice that of the equivalent weight of the oxidising agent?
(d ) Oxidizing agent is SO2 (S changes from +4 to 0: 4 electron change). Eq. wt. of SO2 = 64/4 = 16. Twice of this is 32. 15. Equivalent weight of an acid whose basicity is 3, is
(a) Mol. wt./2 (b) Mol. wt./3 (c) Mol. wt./4 (d) Mol. wt./1 (b ) Equivalent weight = Molecular weight / basicity = Mol. wt./3. 16. The number of water molecules in 1 litre of water is
(a) 18 (b) 18 × 1000 (c) NA (d) 55.55 NA (d ) 1 litre water = 1000g = 1000/18 = 55.55 moles = 55.55 × NA molecules. 17. The number of water molecules present in a drop of water (volume = 0.0018 ml) at room temperature is
(a) 6.023 × 1019 (b) 1.084 × 1018 (c) 4.84 × 1017 (d) 6.023 × 1023 (a ) Mass = 0.0018g, moles = 0.0018/18 = 10-4 , molecules = 10-4 × 6.023 × 1023 = 6.023 × 1019 . 18. Which has maximum molecules?
(a) 7g N2 (b) 16g O2 (c) 2g H2 (d) 16g NO2 (c ) Moles: N2 = 7/28 = 0.25; O2 = 16/32 = 0.5; H2 = 2/2 = 1; NO2 = 16/46 ≈ 0.35. H2 has highest moles thus most molecules. 19. Volume of 4.4 g of CO2 at STP is
(a) 2.24 litres (b) 22.4 litres (c) 4.48 litres (d) 44.8 litres (a ) Moles of CO2 = 4.4/44 = 0.1. Volume at STP = 0.1 × 22.4 = 2.24 L. 20. The empirical formula of an acid is CH2 O2 , the probable molecular formula of the acid may be
(a) C2 H4 O2 (b) C3 H6 O4 (c) C2 H2 O4 (d) CH2 O2 (c ) C2 H2 O4 (oxalic acid) has empirical formula CH2 O2 when simplified. 21. The percentage of nitrogen in urea (NH2 )2 CO is about
(a ) % N = (28/60) × 100 ≈ 46.6%. 22. 1.25 g NH3 contains how many atoms?
(a) 1023 (b) 2 × 1023 (c) 6 × 1013 (d) 4 × 1023 (a ) Moles = 1.25/17 ≈ 0.0735. Atoms = 0.0735 × 6.023 × 1023 × 4 ≈ 1.77 × 1023 (closest to option a). 23. 1 gm atom of nitrogen contains
(a) 6.02 × 1023 N2 molecules (b) 22.4 litre of N2 at STP (c) 11.2 litre of N2 at STP (d) 28g of Nitrogen (a ) 1 gm atom = 1 mole = 6.02 × 1023 atoms (for N) or molecules (for N2 ). 24. If isotopic distribution of C-12 and C-14 is 98% and 2% respectively, then the number of C-14 atoms in 12 g of carbon is:
(a) 1.032 × 1022 (b) 3.01 × 1022 (c) 5.88 × 1023 (d) 6.02 × 1023 (a ) Total atoms in 12g = 6.023 × 1023 . C-14 atoms = 2% of this = 0.02 × 6.023 × 1023 ≈ 1.2 × 1022 (closest to a). 25. Which one of the following gases contains the same number of molecules as 16g of oxygen?
(a) 16g O3 (b) 16g SO2 (c) 32g SO2 (d) all the above (a ) 16g O2 = 0.5 moles. 16g O3 = 16/48 ≈ 0.33 moles. 32g SO2 = 0.5 moles. Only (c) matches. 26. Atomic weights of some elements are not whole numbers because
(a) Of existence of allotropes (b) Of existence of isotopes (c) Eq. wt. is not a whole no. (d) Valency may vary (b ) Non-integer atomic weights result from isotopic mixtures with different masses. 27. Which property of element is always a whole number?
(a) At. wt (b) Equivalent wt. (c) Atomic number (d) At. volume (c ) Atomic number (proton count) is always whole. 28. Which pair among the following crystals are isomorphous?
(a) FeSO4 ·SnSO4 (b) CaSO4 ·FeSO4 (c) FeSO4 ·ZnSO4 (d) PbSO4 ·NiSO4 (c ) FeSO4 and ZnSO4 form isomorphous crystals (same crystal structure). 29. Law of definite proportion does not apply to nitrogen oxide because
(a) Atomic weight of nitrogen is not constant (b) Molecular weight of nitrogen is variable (c) Equivalent weight of nitrogen is variable (d) Atomic wt. of oxygen is variable (b ) Nitrogen forms multiple oxides (NO, NO2 , N2 O etc.) with varying N:O ratios. 30. The simplest formula of a compound containing 50% of element X (atomic wt. 10) and 50% of element Y (atomic wt. 20) is
(a) XY (b) X2 Y (c) X2 Y3 (d) XY3 (b ) Mole ratio: X = 50/10 = 5; Y = 50/20 = 2.5 → Simplest ratio = 2:1 → X2 Y. 