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Solutions — Chemistry STD 12 Science — Question

Bihar BoardEnglish MediumSTD 12 ScienceChemistrySolutions5 Marks
Question
Two elements A and B form compounds having formula AB2 and AB4. When dissolved in 20 g of benzene $\left( C _6 H _6\right), 1 g$ of $AB _2$ 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.
✓

Answer

Freezing point depression :
$\Delta T _{ f }=\frac{ K _{ f } \times w_2 \times 1000}{ M _2 \times w_1}$
Molar mass $\left( M _2\right)=\frac{ K _{ f } \times w_2 \times 1000}{\Delta T_{ f } \times w_1}$
$ M _2= M _{ AB _2}=  \text { Molar mass of } AB _2=\text { ? } $
$ K _{ f }=5.1$ $K$ $kg$ $mol ^{-1} $
$ w_2=1 g, w_1=20 g, \Delta T _{ f }=2.3 K $
$ M _{ AB _2}=\frac{5.1 \times 1 \times 1000}{2.3 \times 20}=110.869 $
$ M _{ AB _2}=110.87$ $g mol ^{-1}$
Similarly, molar mass of AB4
$ M _{ AB _4}  =\frac{ K _{ f } \times w_2 \times 1000}{\Delta T_{ f } \times w_1} $
$\Delta T_{ f }  =1.3 K $
$M _{ AB _4}  =\frac{5.1 \times 1 \times 1000}{1.3 \times 20}=196.15 $
$ =196.15$ $g mol ^{-1}$
Let x and y be atomic masses of A and B respectively, Then
$M _{ AB _2}=x+2 y$
$110.87=x+2 y \quad \quad \ldots \ldots(1)$
$M _{ AB _4}=x+4 y$
$196.15=x+4 y$$\quad \quad \ldots \ldots(2)$
On substracting equation (1) from equation (2),
$ 196.15-110.87  =2 y $
$85.28  =2 y $
$y  =42.64 u$
On putting value of y in equation (1),
$ 110.87  =x+2 \times 42.64 $
$110.87  =x+85.28 $
$x  =110.87-85.28 $
$x  =25.59 u$
Hence, atomic mass of A = 25.59 u
atomic mass of B = 42.64 u

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