### JEE Main On 16 April 2018 Question 28

##### Question: The mass of non-volatile, non-elctrolyte solute (molar mass $ 50gmo{l^{-1}} $ ) needed to be dissolved in114g octane to reduce its vapour pressure to 75%, is ___________. [JEE Main 16-4-2018]

#### Options:

A) 150g

B) 75g

C) 37.5g

D) 50g

## Show Answer

#### Answer:

Correct Answer: A

#### Solution:

Octane has molar mass of 114 g/mol.

$ \frac{\Delta P}{P}=\frac{\frac{W _2}{M _2}}{\frac{W _2}{M _2}+\frac{W _1}{M _1}} $ $ \frac{75}{100}=\frac{\frac{W _2}{50g/mol}}{\frac{W _2}{50g/mol}+\frac{114g}{114g/mol}} $

$ 0.75=\frac{\frac{W _2}{50}}{\frac{W _2}{50}+1} $ $ \frac{W _2}{50}+1=\frac{W _2}{50\times 0.75} $ $ W _2=150,g $ $ W _2=150,g $

Note: $ W _2 $ and $ M _2 $ are mass and molar mass of solute $ W _1 $ and $ M _1 $ and are mass and molar mass of octane.