The vapour pressure lowering caused by addition of 100 g of sucrose (molecular mass = 342) to 1000 g of wate,if the vapour pressure of pure water at `25^(@)C` is 23.8mm Hg,is

Options

A. 1.25 mm Hg
B. 0.125 mm Hg
C. 1.15 mm Hg
D. 00.12 mm Hg

Answer

B

Explanation

Given molecular mass of sucrose = 342
Moles of sucrose = `(100)/(342)=0.292` mole
Moles of water `N=(1000)/(18)=55.5` moles and
Vapour pressure of pure water `P^(o)=23.8` mmHg
According to Raoult.s law
`(DeltaP)/(P^(o))=(n)/(n+N)implies(DeltaP)/(23.8)=(0.292)/(0.292+55.5)`
`DeltaP=(23.8xx0.292)/(55.792)=0.125mmHg`