Vapour pressure of pure water at `298 K ` is `23.8 mm Hg`. `50g` of urea `(NH_(2)CONH_(2))` is dissolved in `850g` of water. Calculate the vapour pressure of water for this solution and its relative lowering.

Options

A. 23.8 mm Hg and 0.16
B. 25.4 mm Hg and 0.02
C. 30.2 mm Hg and 0.020
D. 23.4 mm Hg and 0.017

Answer

D

Explanation

Given ,`p^@`=23.8 mm Hg
`w_2`=50 g, `M_2`(urea)=`60 g mol^(-1),p_s=?,(p^@-p_s)/p^@`=?
`w_1`=850 g, `M_1` (water)=`18 g mol^(-1)`
`therefore n_2=50/60=0.83 , n_1=850/18=47.22`
Applying Raoult's law , `(p^@-p_s)/p^@=n_2/(n_1+n_2)`
or, `(p^@-p_s)/p^@=0.83/(47.22+0.83)=0.83/48.05`=0.017
Thus, relative lowering of vapour pressure =0.017
Again, `Deltap/p^@`=0.017
`therefore Deltap`=0.017 x 23.8
`p^@-p_s` = 0.017 x 23.8
`p_s`=23.8-0.017 x 23.8
`p_s`=23.4 mm Hg
Thus, vapour pressure of water in the solution =23.4 mm Hg