How would you determine molar mass from relative lowering of vapour pressure.

Explanation

(i) The measureement of relative lowering of vapour pressure can be used to determine the molar mass of a non-volatile solute.
(ii) A known mass of the solute is dissolved in a known quantity of solvent. The relative lowering of vapour pressure is measured experimentally.
(iii) According to Raoult.s law, the relative lowering of vapour pressure is
`(P_("solvent")^(0) -P_("solution"))=x _(B)`
`W _(A) = weight of solvent, `W_(B)=` weight of solute
`M _(A)=` Molar mass of solvent,` M_(B)=` molar mass of solute
`therefore x _(B) = (n_(B))/(n _(A)+ n _(B))`
where `n _(A)=` number of moles of solvent, `n _(B)=` number of moles of solute.
For dilute solution, `n _(A) gt gt n _(B), n _(A) + n _(B) ~~ n _(A).`
Then,` x _(B) = (n_(B))/(n _(A))`
Number of moles of solvent and solute are
`n _(A) = (W_(A))/(M_(A)) , n _(B) = (W_(B))/( M _(B))`
`therefore x _(B) = ((W _(B))/(M _(B)))/((W _(A))/(M _(A)))`
Thus, relative lowering of vapour presure ((W _(B))/(M _(B)))/((W _(A))/(M _(A)))`
Relative lowering of vapour pressure ` = (P^(@)-P)/( P^(@))`
`(P^(@) -P)/(P^(@)) = (W_(B) xxM _(A))/(W _(A) xx M _(B))`
From the above equation, molar mass of the folute `M_(B)` can be calculated using the knonw values of `W_(A), W_(B), M_(A)` and the measured relative lowering of vapour pressure.