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

Mass of solute `(W_(B))=5.8 g`.
Mass of solvent i.e., `CS_(2)` is `(W_(A))=100g`
Vapour pressure of solution `(rho)=190` mm of Hg.
Vapour pressure of pure solvent `(P_(A))=195` mm of Hg
Molar mass of pure solvent `CS_(2)(M_(A))=76 gmol^(-1)`.
`M_(B)(p_(A)-p)/(p_(A)^(0))=(W_(B)M_(A))/(M_(B)W_(A))`
`(195-190)/(195)=(5.8xx76)/(M_(B)xx100)`
`(5)/(195)=(5.8xx76)/(M_(B)xx100)`
`M_(B)=(5.8xx76xx195)/(5xx100)=1719 gmol^(-1)`