# QuestionAsked in: JEE Main-2018The mass of a non-volatile, non-electrolyte solute (molar mass=50 g ) needed to be dissolved in 114 g octane to reduce its vapour pressure to 75%, is :A.37.5 gB.75 gC.150 gD.12.5 g

Let massof non-volatile,non-electrolyte solute is w.

molar mass of solute , m = 50 g/mol

mass of solvent , W= mass of octane

=114g

molar mass of solvent , M = 114g/mol

first of all, find mole fraction of solute ;

e.g., mole fraction of solute = mole of solute/(mole of solute + mole of solvent )

= {w/50}/{w/50 + 114/114}

= w/(w + 50)

from lovering of vapor pressure,

?P/P = mole fraction of solute

According to the  to question, vapor pressure is reduced by 75%

then, ?P/P = 0.75

now, 0.75 = w/(w + 50)

or, 3/4 = w/(w + 50)

or, 3w + 150 = 4w

or , w = 150g

hence, mass of non - volatile solute is 150g

## Most Viewed Questions

### Preparation Products

##### Knockout JEE Main (Six Month Subscription)

- AI Coach Study Modules, - Unlimited Mock Tests, - Study Improvement Plan.

₹ 9999/- ₹ 8499/-
##### Knockout JEE Main (Nine Month Subscription)

- AI Coach Study Modules, - Unlimited Mock Tests, - Study Improvement Plan.

₹ 13999/- ₹ 12499/-
##### Test Series JEE Main 2024

Chapter/Subject/Full Mock Tests for JEE Main, Personalized Performance Report, Weakness Sheet, Complete Answer Key,.

₹ 7999/- ₹ 4999/-
##### JEE Main Rank Booster 2023

Booster and Kadha Video Lectures, Unlimited Full Mock Test, Adaptive Time Table, Faculty Support.

₹ 9999/- ₹ 6999/-