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 of the solute are present in 100 g of the solution
`"i.e, Mass of solute "(w_(2))=5g" , Mass of solution = 100 g"`
`therefore" Mass of solvent "(w_(1))=100-5=95g`
Further, as the solution is aqueous, it means that the solvent is water and we know that
Vapour pressure of pure water at 373 K `(p^(@))="760 mm"`
Vapour pressure of the solution at 373 K `(p_(s))="745 mm (Given)"`
Molar mass of solvent (water), `M_(1)="18 g mol"^(-1)`
Molar mass of solute, `M_(2)=" To be calculated"`
Using the formula for dilute solution, viz.,
`" "(p^(@)-p_(s))/(p^(@))=(n_(2))/(n_(1))=(w_(2)//M_(2))/(w_(1)//M_(1))=(w_(2)M_(1))/(w_(1)M_(2))`, we get
`(760-745)/(760)=(5gxx"18 g mol"^(-1))/(95 gxxM_(2))`
`"or "M_(2)=(5xx18xx760)/(15xx95)"g mol"^(-1)="48 g mol"^(-1)`
Assuming that the solution is not dilute, we apply the formula
`(p^(@)-p_(s))/(p^(@))=(n_(2))/(n_(1)+n_(2))=(w_(2)//M_(2))/(w_(1)//M_(1)+w_(2)//M_(2))`
`therefore" "(760-745)/(760)=(5//M_(2))/(95//18+5//M_(2))`
`"or "(95)/(18)+(5)/(M_(2))=(5)/(M_(2))xx(760)/(15)=(760)/(3M_(2))`
`"or "(760)/(3M_(2))-(5)/(M_(2))=(95)/(18)`
`"or "(1)/(M_(2))=((760)/(3)-5)=(95)/(18)`
`"or "(1)/(M_(2))xx(745)/(3)=(95)/(18)`
`"or "M_(2)=(745)/(3)xx(18)/(95)="47.05 g mol"^(-1)`
Note. Usually when the solution is less than `5%`, it is taken as dilute and the approximate formula is employed.
Alternatively, applying the formula, `(p^(@)-p_(s))/(p_(s))=(n_(2))/(n_(1))=(w_(2)//M_(2))/(w_(1)//M_(1))=(w_(2)M_(1))/(w_(1)M_(2))`
`("760 mm"-"745 mm")/("745 mm") or (15)/(745)=(5gxx"18 g mol"^(-1))/(95 g xxM_(2)) or M_(2)="47.05 g mol"^(-1)`.