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.

Explanation

Here, `p^(@)="17.535 mm," w_(2)=25g, w_(1)=450g`
For solute `("glucose, "C_(6)H_(12)O_(6)),M_(2)="180 g mol"^(-1)," For solvent "(H_(2)O),M_(1)="18 g mol"^(-1)`
`"Applying Raout's law, "=(p^(@)-p_(s))/(p^(@))=(n_(2))/(n_(1)+n_(2))or(p^(@)-p_(s))/(p_(s))=(n_(2))/(n_(1))=(w_(2)//M_(2))/(w_(1)//M_(1))or(p^(@))/(p_(s))-1=(w_(2)M_(1))/(w_(1)M_(2))`
Substituting the given values, we get
`(17.535)/(p_(s))-1=(25xx18)/(450xx180)=(250)/(4500) or(17.535)/(p_(s))=1+(25)/(4500)=(4525)/(4500)orp_(s)=17.353xx(4500)/(4525)="17.44 mm"`