Question

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

• A. $13.88 \times 10^{-1}$
• B. $13.88 \times 10^{-2}$
• C. $13.88$
• D. $13.88 \times 10^{-3}$

Answer 1

Answer: (C)

$X_{\text{solvent}} = 0.8 $
If $ n_{T} =1 $
$n_{\text{Solvent}} = 0.8 $
$n_{\text{Solute}} =0.2 $
molality $= \frac{0.2}{\frac{0.8\times18}{1000}} = 13.88 $