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The vapour pressure of water at 293K is...

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

Text Solution

Verified by Experts

`P^(@)=17*535mm`
Molar mass of glucose `=180" g "mol^(-1)`
Molar mass of water `=18" g "mol^(-1)`
According to Raoult's law,
`(P^(@)-P_(S))/(P^(@))=(n_(2))/(n_(1)+n_(2))=(n_(2))/(n_(1))=(W_(2)//M_(2))/(W_(1)//M_(1))`
`1-(P_(S))/(P^(@))=(25//180)/(450//18)=(25xx18)/(180xx450)`
`1-(P_(S))/(P^(@))=(450)/(81000),1-(450)/(81000)=(P_(S))/(P^(@))`
`1-0.0055=(P_(S))/(17*535`
`0.9945=(P_(S))/(17*535)`
`thereforeP_(S)=0.9945xx17.535=17.44` mm Hg.
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