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Chemistry Question on Mole concept and Molar Masses

The mole fraction of a solvent in aqueous solution of a solute is 0.8. The molality (in mol kg−1kg^{-1}) of the aqueous solution is

A

13.88×10−113.88 \times 10^{-1}

B

13.88×10−213.88 \times 10^{-2}

C

13.8813.88

D

13.88×10−313.88 \times 10^{-3}

Answer

13.8813.88

Explanation

Solution

Xsolvent=0.8X_{\text{solvent}} = 0.8
If nT=1n_{T} =1
nSolvent=0.8n_{\text{Solvent}} = 0.8
nSolute=0.2n_{\text{Solute}} =0.2
molality =0.20.8×181000=13.88= \frac{0.2}{\frac{0.8\times18}{1000}} = 13.88