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

Answer

A

Explanation

`W_(B) = 0.5g`
`M_(B) = 65` amu
`W_(A) = "vol" xx "density" = 100 xx 1.58 = 158g`
`M_(A)(C CI_(4)) = 12 +35.5 xx 4 = 154`
`(P^(@)-P_(s))/(P^(@)) = (W_(B))/(M_(B)).(M_(A))/(W_(A))`
`P^(@) - P_(S) = (W_(B)M_(A))/(M_(B)W_(A)) xx P^(@)`
`P_(S) = P^(@) - (W_(B)M_(A))/(M_(B)W_(A)) xx P^(@)`
`= 143 - (0.5 xx 154)/(65 xx 158) xx 143`
`= 143 - 1.03 = 141.97mm`