A solution containing`0.5g` of a non-volatile solute in `0.2dm^(3)` of the solution exerts an osmotic pressure of `44.44 kPa` at `300K`. Thus, molar mass of the solute is

Options

A. `150gmol^(-1)`
B. `300gmol^(-1)`
C. `140gmol^(-1)`
D. `168gmol^(-1)`

Answer

C

Explanation

`pi=(w)/(mV)RT`
Given,`pi=66.66kaPa`
`R=8.3143JKmol^(-1)=8.3143kPadm^(3)K^(-1)mol^(-1)`
`w=0.4g.m=?,V=0.1dm^(3),T=300K`
`:.m=(wRT)/(underset(')AV)=(0.5xx8.3143xx300)/(44.44xx0.1)`
`=140.4~~140.0gmol^(-1)`