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))= 5.8g`
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)^(0))= 195` mm of Hg
Molar mass of pure solvent `CS_(2)(M_(A))=76" gmol"^(-1)`
`"M"_(B) (p_(A)^(0)-p)/(p_(A)^(0))= (W_(B)MA)/(M_(B)W_(A))`
`(195-190)/(195)= (5.8xx76)/(M_(B)xx100)`
`(5)/(195)= (5.8 xx 76)/(M_(B)xx100)`
`M_(B)=(5.8xx76xx195)/(5xx100)=171.9" gmol"^(-1)`