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

Answer

`P_(s)=17.44mm Hg`

Explanation

Given, `p^(@)=17.535mm, W_(2)=25g,W_(1)=450g`
Molecular mass glucose `=180g mol^(-1)`
Using Raoult's law `:`
`(p^(@)-p_(S))/(p^(@))=(n_(2))/(n_(1)+n_(2))`
or `(p^(@)-p_(S))/(p_(S))=(n_(2))/(n_(1))=(W_(2)//Mw_(2))/(W_(1)//Mw_(1))`
or `(p^(@))/(p_(S))-1=(W_(2)xxMw_(1))/(W_(1)xxMw_(2))`
Substituting the given values,
`(17.535)/(p_(S))-1=(25xx18)/(450xx180)=(25)/(4500)`
or `(17.535)/(p_(S))=1+(25)/(4500)=(4525)/(4500)=1.005`
`p_(S)=17.535xx1.005=17.44mm`