`18g` glucose `(C_(6)H_(12)O_(6))` is added to `178.2g` water. The vapour pressure of water (in torr) for this aqueous solution is:

Options

A. `759.0`
B. `7.6`
C. `76.0`
D. `752.4`

Answer

D

Explanation

From Roult.s law
`(P_(O)-P_(S))/(P_(O))=X_("solute")`
`therefore (P_(O)-P_(S))/(P_(O))=(n_("solute"))/(n_("solute")+N_("solvent"))`
`rArr (P_(O)-P_(S))/(P_(O))=(0.1)/((178.2)/(18)+0.1)`
`because P_(o)=760` torr `(100^(@)C)`
`therefore` Vapour pressure of `H_(2)O` in solution = 752.3 torr