Vapour pressure of `C CL_(4)` at `25^@C` is `143` mmHg 0.05g of a non-volatile solute (mol.wt.=`65`)is dissolved in `100ml C CL_(4)`. find the vapour pressure of the solution (density of `C CL_(4)=158g//cm^2`)

Options

A. `141.9 mmHg`
B. `94.4mmHg`
C. `-1.86^(@)C`
D. `0.93^(@)C`

Answer

A

Explanation

Wt. of `CCl_(4)=(153.8)/(154)approx1`
moles of solute `=(0.5)/(65)=0.00769`
Now `(P_(0)-P_(S))/(P_(S))=(n)/(N)`
or `(143-P_(S))/(P_(S))=(0.00769)/(1)`
so `P_(s)=141.9mm Hg`.