Question

Molarity (M) of an aqueous solution containing $x \, \text{g}$ of anhyd. CuSO$_4$ in 500 mL solution at 32$^\circ$C is $2 \times 10^{-1} \, \text{M}$. Its molality will be ____ $\times 10^{-3} \, \text{m}$ (nearest integer).
[Given density of the solution = 1.25 g/mL.]

Answer 1

Given:

• Molarity ($M$) = 0.2 mol/L
• Volume of solution ($V_{sol}$) = 500 mL = 0.5 L
• Density of solution ($d_{sol}$) = 1.25 g/mL
• Molar mass of CuSO$_4$ = 159.5 g/mol

Step 1: Calculate the mass of the solution

$M_{sol} = V_{sol} \times d_{sol} = 500 \, \text{mL} \times 1.25 \, \text{g/mL} = 625 \, \text{g}.$

Step 2: Calculate the mass of solute

$\text{Mass of solute (CuSO}_4\text{)} = M \times V_{sol} \times \text{Molar mass}.$ $= 0.2 \times 0.5 \times 159.5 = 15.95 \, \text{g}.$

Step 3: Calculate the mass of the solvent

$\text{Mass of solvent} = \text{Mass of solution} - \text{Mass of solute}.$ $= 625 - 15.95 = 609.05 \, \text{g} = 0.60905 \, \text{kg}.$

Step 4: Calculate the molality

$m = \frac{\text{Moles of solute}}{\text{Mass of solvent (in kg)}} = \frac{0.1}{0.60905}.$ $m = 0.164 \, \text{mol/kg} = 164 \times 10^{-3} \, \text{mol/kg}.$

Final Answer

The molality of the solution is:

$0.164 \, \text{mol/kg (or } 164 \times 10^{-3} \, \text{mol/kg)}.$