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Question : 21 of 53
Marks: +1, -0
Two elements A and B form compounds having molecular formula AB2\mathrm{AB}_2 and AB4\mathrm{AB}_4. When dissolved in 20.0 g of benzene (C6H6)(\mathrm{C_6H_6}), 1.0 g AB2\mathrm{AB}_2 lowers the freezing point by 2.3°C whereas 1.0 g of AB4\mathrm{AB}_4 lowers the freezing point by 1.3 °C. The molal depression constant for benzene is 5.1 K kg mol −1^{-1}. Calculate atomic mass of A and B.
Solution:  
For AB2,ΔTf=Kf×w2×1000M2×w1\mathrm{AB}_2, \Delta T_f = K_f \times \frac{w_2 \times 1000}{M_2 \times w_1} or, 2.3=5.1×1×1000M2×202.3 = 5.1 \times \frac{1 \times 1000}{M_2 \times 20}
∴    M2=50×5.12.3=110.86  g  mol−1\therefore\;\; M_2 = \frac{50 \times 5.1}{2.3} = 110.86 \; \text{g} \; \text{mol}^{-1}
Similarly for AB4,1.3=5.1×1×1000M2×20\mathrm{AB}_4, 1.3 = 5.1 \times \frac{1 \times 1000}{M_2 \times 20}
∴    M2=50×5.11.3=196.15  g  mol−1\therefore\;\; M_2 = \frac{50 \times 5.1}{1.3} = 196.15 \; \text{g} \; \text{mol}^{-1}
Now, molecular weight of AB2=110.86,\mathrm{AB}_2 = 110.86, molecular weight of AB4=196.15\mathrm{AB}_4 = 196.15
AB4=A+4B=196.15…(i)\mathrm{AB}_4 = A + 4B = 196.15 \qquad \ldots (i)
AB2=A+2B=110.86…(ii)\mathrm{AB}_2 = A + 2B = 110.86 \qquad \ldots (ii)
(i) −(-( ii) gives 2B=85.292B = 85.29
∴    B=42.645  u\therefore\;\; B = 42.645 \; \text{u}
Putting the value of B in equation (ii),
A+2×42.645=110.86A + 2 \times 42.645 = 110.86 or, A=110.86−85.29=25.57  uA = 110.86 - 85.29 = 25.57 \; \text{u}
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