Question

The mole fraction of a solvent in aqueous solution of the solute is 0.8. The molality (in mol kg^-1) of the aqueous solution is: 
1. 13.88 × 10^-2
2. 13.88 × 10^-1
3. 13.88
4. 13.88 × 10^-3

Answer 1

Correct Answer - Option 3 : 13.88

Calculation:

Let, 1 mole be present in the solution,

Solvent = 0.8 mol and

Solute = 0.2 mol

Molar Mass of water = 18 g/kg

= 0.8 × 18 g =14.4

\({\rm{Molality}} = \frac{{{\rm{Moles\;of\;Solute}}}}{{{\rm{Kilogram\;of\;solvent}}}} = {\rm{\;}}\frac{{0.2}}{{\frac{{0.8 \times 18}}{{1000}}}}\)

\(\therefore {\rm{Molality}} = \frac{{0.2\; \times {\rm{\;}}1000}}{{0.8\; \times {\rm{\;}}18}}\)

\(= \frac{{200}}{{14.4}} \approx 13.88\)