An aqueous solution freezes at 272.4 K while pure water freezes at 273 K. Given `K_(f)=1.86 K kg "mol"^(-1)`,`K_(b)=0.512 K kg "mol"^(-1)` and vapour pressure of water at 298 K = 23.756 mm Hg. Determine the following.
Lowering in vapour pressure at 298 K is

Options

A. `0.13`
B. `0.15`
C. `0.16`
D. `0.1378`

Answer

D

Explanation

Lowering in vapour pressure
According to Raoult's law
`(P^(@)-P)/P^(@)=chi_(B)`
Now, `chi_(B)=n_(B)/(n_(A) xx n_(B))=n_(B)/n_(A)` (for dilute solution)
=`n_(B)/(W_(B)) xx Mw_(A)` ….(i)
and m=`n_(B) xx 1000/W_(A)` ….(ii)
Dividing Eq. (i) by Eq. (ii), we get
`chi_(B)/m=(Mw_(A))/1000 rArr chi_(B)=(m xx Mw_(A))/1000`
`:. chi_(B)=(0.322 xx 18)/(1000)=0.00580`
Now `(P^(@)-P)/P^(@)=0.00580`
or `P^(@)-P=P^(@) xx 0.00582 = 23.756 xx 0.00580`
`=0.1378 mm`
`:.` Lowering in vapour pressure =0.1378 mm