The mole fraction of a solvent in aqueous solution of a solute is 0.8. The molality `("in mol kg"^(-1))` of the aqueous solution is:

Options

A. `13.38 xx 10^(-1)`
B. `13.88 xx 10^(-2)`
C. 13.88
D. `13.88 xx 10^(-3)`

Answer

C

Explanation

x(solvent) = 0.8
If total number of moles =1
x(solvent) =0.8, n(solute) `=1-0.8=0.2`
Molality `=(0.2)/(0.8 xx 18) xx 1000 = 13.88`