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Vapour pressure of water at 293 K is 17....

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 dissoved in 450 g of water.

Text Solution

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According to Raoult's Law
`(P_(A)^(@)-P_(S)^(@))/P_(S)=n_(B)/n_(A)or P_(A)^(@)/P_(S)-1=n_(B)/n_(A)`
`P_(A)^(@)/P_(S)=1+n_(B)/n_(A)=1+W_(B)/M_(B)xxM_(A)/W_(A)`
`W_(B)=256 g, W_(A)=450g , M_(B)=180 g mol^(-1),`
` M_(A)=18 g mol^(-1), P_(A)^(@)=17.535 mm`
`P_(A)^(@)/P_(S)=1+((25g)xx(18g mol^(-1)))/((180f mol^(-1))xx(450g))=1+0.0055=1.0055`
`P_(S)("V.P of water in solution")=((17.55mm))/((1.0055))=17.44 mm`
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