An aqueous solution of `2` per cent `(wt.//wt)` non-volatile solute exerts a pressure of `1.004` bar at the boiling point of the solvent. What is the molecular mass of the solute?

Options

A. `"23.4 g mol"^(-1)`
B. `"41.35 g mol"^(-1)`
C. `"10 g mol"^(-1)`
D. `"20.8 g mol"^(-1)`

Answer

B

Explanation

Vapour pressure of pure water at boiling point =1 atm = 1.013 bar
Vapour pressure of solution `(p_s)`=1.004 bar
Let mass of solution = 100 g
Mass of solute=2 g
Mass of solvent = 100-2= 98 g
`(p^@-p_s)/p^@=n_2/(n_1+n_2)=n_2/n_1 =(W_2//M_2)/(W_1//M_1) ( because n_2 lt lt lt n_1)`
`(1.013-1.004)/1.013=2/M_2xx18/98`
or `M_2=(2xx18)/98xx(1.013)/0.009 = 41.35 g mol^(-1)`