1.

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

Answer»

`759.0`
`7.6`
`76.0`
`752.4`

Solution :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


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