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)^(@)-P)/P_(1)^(@)=n_(2)/n_(1), (143-p)/143 = (0.5//65)/(158//154) :. p = 141.93`