An aqueous solution of glucose is made by dissolving 10 g of glucose in 90 g water at 303 K. If the V.P. of pure water at 303 K be 32.8 mm Hg, what would be V.P. fo the solution ?

Explanation

According to Raoult.a law, vapour pressure of the solution
`p_(A) = p_(A)^(@) x_(A)`
where `p_(A)^(@)` = vapour pressure of pure water and `x_(A)` is the mole fraction of water .
Since solute in non- volatile .
`p_(a)^(@)` = 32.8 mm Hg
Moles of water `= (90)/(18) = 5.0`
Moles of glucose `= (10)/(180) = 0.0556`
Moles fraction of water , `x_(A) = (5.0)/(5.0 + 0.0556) = 0.989`
Vapour pressure of solution ` = 32.8 xx 0.989`
= 32.44 mm Hg