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Standard X

Chemistry

Mole Concept

Two elements ...

Question

Two elements form compounds having molecular formula $A B_{2}$ and $A B_{4}$ . When dissolved in 20 g of benzene, 1.0 g of $A B_{4}$ lowers it by 1.3 K. The molar depression constant for benzene is 5.1 K kg $m o l^{- 1}$ . Calculate the atomic mass of $A$ and $B$ .

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Solution

Given, $K_{f} = 5.1$ K kg ${m o l}^{- 1}$
$W_{1} = 20$ $W_{2} = 1$
We know, ${Δ T}_{f} = K_{f} m$
$m = \frac{W_{2} \times 1000}{M_{2} \times W_{1}}$
$∴$ $M_{2} = \frac{1000 \times K_{f} \times W_{2}}{W_{1} \times {Δ T}_{f}}$
for ${A B}_{4}, {Δ T}_{f} = 1.3$ K
$M_{2} = \frac{1000 \times 5.1 \times 1}{1.3 \times 20}$
$M_{2} = 196.15$ g ${m o l}^{- 1}$
let mass of $A = x g$
mass of $B = y g$
for ${A B}_{4}, x + 4 y = 196.15 g \to (1)$
for ${A B}_{2}, x + 2 y = M_{2}$
$M_{2} = \frac{1000 \times 5.1 \times 1}{2.3 \times 20}$
$M_{2} = 110.87 g = x + 2 y \to (2)$
solving $(1)$ and $(2)$
$x + 4 y = 196.15 g$
$x + 2 y = 11.87 g$
mass of $A = x = 25.5 g$
mass of $B = y = 42.6 g$

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