At room temperature, a dilute solution of urea is prepared by dissolving 0.60 g of urea in 360 g of water. If the vapour pressure of pure water at this temperature is 35 mm Hg, lowering of vapour pressure will be: (molar mass of urea `=60 g "mol"^(-1)`)

Answer

0.017

Explanation

Relative loweing of vapour pressure, is given by, `(p^(0)-p)/p^(0) =x_(A) =n_(A)/(n_(A) + n_(B)) =n_(A)/n_(B)`
Given, `p^(0)= 35` mm Hg, `n_("urea") = 0.60/60`
`n_("water") = 360/18`
`(p^(0)-p)/35 =(0.6 xx 18)/(60 xx 360) = 1/2000`
`Deltap = p^(0) -p=0.017`