5.8 g of non - volatile, non - electrolyte solute was dissolved in 100 g of carbon disuiphide `(CS_(2))`. The vapour pressure of the solution was found to be 190 mm of Hg. Calculate molar mass of the solute. Given : Vapour of pure `CS_(2)` is 195 mm of Hg and molar mass of `CS_(2)` is `76g//mol`.

Explanation

`w_(B)=5.8g`
`w_(A)=100g`
`p=190mm-Hg`
`M_(B)=?`
`P_(CS_(2))^(0)="195 mm Hg"`
`M_(A)="76 gmol"^(-1)`
`((P^(@)-P)/(P^(@)))=(w_(B)M_(A))/(w_(A).M_(B))`
`((195-190)/(195))=(5.8xx76)/(100xxM_(B))`
`M_(B)=(4.408)/(0.0256)`
`M_(B)="172.18 g mol"^(-1)`