Question

If 4 g of $$\mathrm{NaOH}$$ dissolves in 36 g of $$\mathrm{H}{\phantom{A}}_2\mathrm{O}$$, calculate the mole fraction of each component in the solution. Also, determine the molarity of solution (specific gravity of solution is 1g mL^–1).

Answer 1

Molar mass of $$\mathrm{NaOH}$$ = 40 g/mole

Molar mass of water = 18 g/ mole

Mass of $$\mathrm{NaOH}$$ = 4 g

Number of moles of NaOH = ${\displaystyle \frac{4 g}{40 g}}$ = 0.1 mol

Number of moles of H₂O = ${\displaystyle \frac{36 g}{18 g}}$ = 2 mol

Mole fraction of water = ${\displaystyle \frac{26 g}{\text{Number of moles of NaOH}+\text{Number of moles of} H_{2}O}}$

Mole fraction of NaOH = ${\displaystyle \frac{2}{0.1+2}=\frac{2}{2.1}}$ = 0.95

Number of moles of ${\displaystyle \frac{2}{0.1}=\frac{0.1}{2.1}}$ = 0.047

Number of moles of $$\mathrm{NaOH}$$

Mass of solution = Mass of solute + Mass of solvent

Mass of $$\mathrm{NaOH}$$ + Mass of water = 4 g + 36 g = 40 g

Specific gravity of solution = 1 g/ml

1 litre = 1000 ml volume of solution = 40 ml

∴ 40 ml = 0.04 litre

Molarity = ${\displaystyle \frac{\text{Number of moles of solute}}{\text{Volume in litre}}=\frac{0.1 \text{mol}}{0.04 \text{litre}}}$ = 2.5 M