When a liquid that is immiscible with water was steam distilled at `952^(@)C` at a total pressure of `748` torr, the distillate contained `1.25g` of the liquid per gram of water. The vapour pressure of water is `648` torr at `95.2^(@)C`. What is the molar mass of liquid?

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)`.