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Solutions — Chemistry STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceChemistrySolutions3 Marks
Question
Two elements A and B form compounds having formula AB2 and AB4. When dissolved in 20 g of benzene (C6H6), 1 g of AB2 lowers the freezing point by 2.3 K whereas 1.0 g of AB4 lowers it by 1.3 K. The molar depression constant for benzene is 5.1 K kg mol–1. Calculate atomic masses of A and B.
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Answer

$\text{Using the relation},\text{ M}_2=\frac{1000\times\text{k}_\text{f}\times\text{w}_2}{\text{w}_1\times\Delta\text{T}_\text{f}}$

$\therefore\ \text{M}_{\text{AB}_2}=\frac{1000\times5.1\times1}{20\times2.3}=110.87\text{ g mol}^{-1}$

$\text{M}_{\text{AB}_4}=\frac{1000\times5.1\times1}{20\times1.3}=196.15\text{ g mol}^{-1}$

Let the atomic masses of A and B are 'p' and 'q' respectively.

Then molar mass of

AB2 = p + 2q = 110.87 g mol-1 .... (i)

And molar mass of

AB4 = p + 4q = 196.15 g mol-1 .... (ii)

Substracting equation (ii) from equation(i), we

get 2q = 85.28 ⇒ q = 42.64

Putting q = 42.64 in equ. (i), we get

p = 110.87 - 85.28

p = 25.59

Thus, atomic mass of A = 25.59 g mol-1 and atomic mass of B = 42.64 g mol-1

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