Question

Calculate the molality of 12.5% w/w Sulphuric acid?

Answer 1

Molality

• Molality is defined as the number of moles of solute per kg of solvent.
  Molality = $\frac{\mathrm{Number}\mathrm{of}\mathrm{moles}\mathrm{of}\mathrm{solute}}{\mathrm{Mass}\mathrm{of}\mathrm{solvent}\mathrm{in}\mathrm{kg}}$

Calculation of Molality

Step 1: Determination of the mass of solute and solvent
12.5% w/w means 12.5 g in 100 g of solution.
Weight of solvent = 100 g – 12.5 g
Weight of solvent = 87.5 g.

Step 2: Determination of the number of moles of solute
Molar mass of Sulphuric acid (${\mathrm{H}}_2{\mathrm{SO}}_4$) = 98g/mol
Number of moles of Sulphuric acid (${\mathrm{H}}_2{\mathrm{SO}}_4$) = $\frac{12.5}{98}$
Number of moles of ${\mathrm{H}}_2{\mathrm{SO}}_4$ = 0.127 mol

Step 3: Determination of molality
Molality = $\frac{\mathrm{Number}\mathrm{of}\mathrm{moles}\mathrm{of}\mathrm{solute}}{\mathrm{Mass}\mathrm{of}\mathrm{solvent}\mathrm{in}\mathrm{kg}}$
Molality = $\frac{0.127\mathrm{mol}}{87.5\mathrm{g}\times {\displaystyle \raisebox{1ex}{$1\mathrm{kg}$}\!\left/ \!\raisebox{-1ex}{$1000$}\right.}\mathrm{g}}$
Molality = $\frac{0.127\mathrm{mol}\times 1000}{87.5\mathrm{kg}}$
Molality = 1.45 mol/kg

Therefore, the molality of 12.5 %w/w Sulphuric acid solution is 1.45 m