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

Data: Vapoure pressure of solvent`=P^(0)=195mm`
Vapour pressure of solution`=P=190mm`
`P^(@)-P=195-190=5mm`
Relative lowering of vapour pressure`=(P^(@)-P)/(P)=(5)/(195)=0.02564`
Mass of solution in grams=`W_(2)=5.8g`,
Mass of solvent `(CS_(2))` in grams=100g,
Molar mass of solvent `(CS_(2))=76g" "mol^(-1)`,
* Formula, `M_(2)=(W_(2))/(W_(1))xx(M_(1))/([(P^(@)-P)/(P)])`
Substitution: `M_(2)=(5.8)/(100)xx(76)/(0.02564)`
Answer `M_(2)=171.92g//mol`.