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Standard XII

Chemistry

Osmotic Pressure

A 20 g non vo...

Question

A 20 g non volatile solute X (Molecular weight = 50 g/mol) trimerises, when dissolved in 1 kg benzene. If degree of trimerisation is 1/4, find abnormal molar mass of X?

200

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150

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120

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60

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Solution

The correct option is D $60$
Given,
Degree of dissociation= $1 / 4 = α$ , $n = 3$
$i = 1 - α + α / n$
$= 1 - 0.25 + 0.25 / 3$
$= 0.83$
$i = \frac{Normal molecular mass}{Observed molar mass}$
$\Rightarrow 1.5 = \frac{50}{Observed molar mass}$
$∴$ Observed molar mass= $\frac{50}{0.83} = 60.24$ (Abnormal molar mass)

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A 20 g non volatile solute X (Molecular weight = 50 g/mol) trimerises, when dissolved in 1 kg benzene. If degree of trimerisation is 1/4, find abnormal molar mass of X?

The correct option is D 60Given,Degree of dissociation= 1/4=α, n=3i=1−α+α/n =1−0.25+0.25/3 =0.83i=Normal molecular massObserved molar mass⇒1.5=50Observed molar mass∴ Observed molar mass= 500.83=60.24 (Abnormal molar mass)