A liquid is immiscible in water was steam distilled at `95.2^(@)C` at a pressure of `0.983` atm. What is the mass of the liquid present per gram of water in the distullate. Molar mass of the liquid is `134.3` g/mol and the vapour pressure of water is `0.84` atm. Also, Vapour pressure of pure liquid is `0.143` atm.

Options

A. `134.1 g mol^(-1)`
B. `105.74 g mol^(-1)`
C. `99.65g mol^(-1)`
D. `18g mol^(-1)`

Answer

A

Explanation

`P_("total")=99.652KPa`
`P_("water")=85.140KPa`
`P_("liquid")=(99.652-85.140)Kpa=14.512KPa`
and `(m_(A))/(m_(B))=(1.27g)/(1g)`
we have, `(m_(A))/(m_(B))=(P_(A)M_(A))/(P_(B)M_(B))`
or `M_(A)=(m_(A))/(m_(B))(P_(B)M_(B))/(P_(A))`
`therefore M_(A)=(1.27)((85.140KPaxx18gmol^(-1))/(14.512KPa))~=134.1gmol^(-1)`.