Question

If an aqueous solution of Na2CO3, mole fraction of Na2CO3 is 0.2, then molality of the solution is (given: molar mass of Na2CO3 is 106)

Answer 1

,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,
molaliti=1.3888

Answer 2

Answer:

The molality of the solution is 13.88 m.

Explanation:

Mole fraction of sodium carbonate in solution = $\chi_1=0.2$

Mole fraction of water in solution = $\chi_2=?$

We know that sum of mole fraction is is equal to 1.

$\chi_1+\chi_2=1$

$\chi_2=1-\chi_1=1-0.2=0.8$

This means that 0.2 moles of sodium carbonate is present in 1 mol of solution.

$Molality=\frac{\text{Moles of compound}}{\text{Mass of solvent in kg}}$

Mass of 0.8 moles of water =$0.8 mol\times 18 g/mol=14.4 g=0.0144 kg$

$Molality=\frac{0.2 mol}{0.0144 kg}=15.625 mol/kg=13.88 m$

The molality of the solution is 13.88m.