# The mole fraction of a solvent in aqueous solution of a solute is $0.8.$ The molality $\left ( in\: \: \: mol\: Kg^{-1} \right )$  of the aqueous solution is : Option 1)$13.88\times 10^{-2}$Option 2)$13.88\times 10^{-3}$Option 3)$13.88\times 10^{-1}$  Option 4)$13.88$

Mole Fraction -

It is ratio of moles of solute or moles of solvent to moles of solution.

- wherein

If a substance ‘A’ dissolves in substance ‘B’ and their number of moles are $n_{A}$ and $n_{B}$ respectively; then the mole fractions of A and B are given as

Mole fraction of A = (number of moles of A)/(number of moles of solution ) = $n_{A}$/($n_{A}$+ $n_{B}$)

We have,

$mole\: \: fraction=0.8$

$X_{solvent}=\frac{n_{solvent}}{n_{solute}+n_{solvent}}=\frac{0.8}{1}$

$n_{solute}=0.2mole$

$molality=\frac{0.2}{0.8\times 18}\times 1000$

$=13.88$

Option 1)

$13.88\times 10^{-2}$

Option 2)

$13.88\times 10^{-3}$

Option 3)

$13.88\times 10^{-1}$

Option 4)

$13.88$

### Preparation Products

##### Knockout JEE Main Sept 2020

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 12999/- ₹ 6999/-
##### Rank Booster JEE Main 2020

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 9999/- ₹ 4999/-
##### Test Series JEE Main Sept 2020

Take chapter-wise, subject-wise and Complete syllabus mock tests and get in depth analysis of your test..

₹ 4999/- ₹ 1999/-
##### Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-