`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. 7.6
B. 76
C. 752.4
D. `759.0`

Answer

C

Explanation

`"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
= Mole fraction of water in the solution `xx` Vapour pressure of pure water
`=0.99xx"760 torr = 725.4 torr"`