At 298 K 1.0 g of a non-colatile solute is dissoved in 100 g of acetone (mol mass=58). The vapour pressure of the solution at this temperature is found to be 192.5 mm Hg. Calculate themolar mass of the solute. The vapour pressure of pure acetone at 298 K is found to be 195 mm Hg.

Options

A. 25.24
B. 35.24
C. 50
D. 55.24

Answer

C

Explanation

P = Vapour pressure of solution = 1.92.5mm Hg
`P_(S)` = Vapour pressure of pure solvent = 195mmHg
`chi_(s)` = Mole fraction of solvent (acetone)
`n_(s)` = Moles of solvent (acetone) = `(100g)/(58g)`
= 1.72 mol
n = moles of solute
Total number of moles in solution = 1.72 + n
`P=P_(S)chi_(S)`
`192.5=195xx((1.72)/(1.72+n))`
`(192.5)/(195)=(1.72)/(1.72+n)`
`(0.99)(1.72+n)=1.72`
`1.72+n=(1.72)/(0.99)`
`1.72+n=1.74`
n = 0.02
moles of solute `(n) = ("given mass")/("molar mass")`
`0.02=(1g)/("molar mass")`
molar mass of solute = `(1)/(0.02)`
`=50gmol^(-1)`.