The vapour pressure of a `5%` aqueous solution of a non-volatile organic substance at `373 K`. Is `745 mm`. Calculate the molecular mass of the solute.

Explanation

`5%` aqueous solution of the solute implies that `5 g` of the solute are present in `100 g` of the solution, i.e.,
Weight of solute `(W)_(2)=5 g`
Weight of solution=100 g
`:.` Weight of solvent `(W_(1))=100-5 =95 g`
Further as the solution is aqueous, it means that the solvent is water and we know that vapour pressure of pure water at `393 K =760mm`
Vapour pressure of solvent at `373 K (P_(s))=745 mm` (given)
Molecular mass of solvent (water), `Mw_(1)=18`
Molecular massof solute, `Mw_(2)`= To be calculated
Using the formula for dilute solutions, viz
`(P^(@)-P_(s))/P^(@)=n_(2)/n_(1)=(W_(2)//Mw_(2))/(W_(1)//Mw_(1))`
We get (760-745)/760=(5//Mw_(2))/(95//18) `
or `Mw_(2)=(5 xx 18 xx 760)/(15 xx 95)=48`