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»

7.6
76
752.4
`759.0`

Solution :`"18 g glucose "(C_(6)H_(12)O_(6))=(18)/(180)" MOLE"="0.1 mole"`
`"178.2 g water "(H_(2)O)=(178.2)/(18)" mole"="9.9 moles"`
`THEREFORE"Mole fraction of water in the solution"`
`=(9.9)/(0.1+9.9)=(9.9)/(10)=0.99`
Vapour pressure of water in aqueous solution
= Molefraction of water in the solution `xx` Vapour pressure of pure water
`=0.99xx"760 torr= 725.4 torr"`


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