Vapour pressure of `"CCl"_(4)` at `25^(@)C` is 143 mm Hg . 0.5 g of a non-volatile solute ( molar mass = `65 mol^(-1)`) is dissolved in 100 mL of `"CCl"_(4)` (density = 1.538g `mL^(-1)` ) Vapour pressure of solution is :

Options

A. 141.93 mm
B. 94.39 mm
C. 199.34 mm
D. 143.99 mm

Answer

A

Explanation

`(P_(1)^(o)-P)/P_(1)^(o)=n_(2)/n_(1)`
`(143-P)/143=(0.5//65)/(158//154)`
or P=141.93 mm