Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.

Explanation

Mass of glucose, `w_(2)=25g`
Mass of water, `w_(1)=450 g`
We know that,
Molar mass of glucose `(C_(6)H_(12)O_(6))`
`M_(2)=6xx12+12xx1+6xx16`
`= 180 g mol^(-1)`
Molar mass of water `M_(1)=18 g mol^(-1)`
Then, number of moles of glucose,
`n_(2)=(25)/(180 g mol^(-1))=0.139` mol
And, number of moles of water,
`n_(2)=(450)/(18 g mol^(-1))=25 mol`
We know that,
`(p_(1)^(0)-p_(1))/(p_(1)^(0))=(n_(1))/(n_(2)+n_(1))`
`(17.535-p_(1))/(17.535)=(0.139)/(0.139+25)`
`17.535-p_(1)=(0.139xx17.535)/(25.139)=0.00552`
`17.535-p_(1)=0.087`
`p_(1)=17.44` mm of Hg
Hence, the vapour pressure of water is 17.44 mm of Hg.