Question

Sea Water is found to contain 5.85% NaCl and 9.50% MgCl2 by weight of solution . Calculate its normal boiling point assuming 80% ionisation for NaCl and 50% ionisation for MgCl2

Answer 1

102.3 °C

Answer 2

To calculate the normal boiling point of the sea water solution, we need to consider the colligative properties, specifically the boiling point elevation. Here's a step-by-step breakdown:

1. Calculate the mass of water: Assuming 100 g of solution, the mass of water is $100 - 5.85 - 9.50 = 84.65$ g, which is 0.08465 kg.

2. Calculate the moles of NaCl and MgCl$_2$:

   • Moles of NaCl: $\frac{5.85 \text{ g}}{58.5 \text{ g/mol}} = 0.1 \text{ mol}$
   • Moles of MgCl$_2$: $\frac{9.50 \text{ g}}{95.0 \text{ g/mol}} = 0.1 \text{ mol}$

3. Calculate the van't Hoff factor ($i$) for each solute:

   • For NaCl (dissociates into 2 ions): $i_{\text{NaCl}} = 1 + 0.80(2-1) = 1.80$
   • For MgCl$_2$ (dissociates into 3 ions): $i_{\text{MgCl}_2} = 1 + 0.50(3-1) = 2.00$

4. Calculate the effective moles of particles:

   • Effective moles from NaCl: $0.1 \text{ mol} \times 1.80 = 0.18 \text{ mol}$
   • Effective moles from MgCl$_2$: $0.1 \text{ mol} \times 2.00 = 0.20 \text{ mol}$

5. Calculate the total effective moles of solute particles:

   • Total effective moles: $0.18 \text{ mol} + 0.20 \text{ mol} = 0.38 \text{ mol}$

6. Calculate the molality ($m$) of the solution:

   • Molality: $\frac{0.38 \text{ mol}}{0.08465 \text{ kg}} \approx 4.489 \text{ mol/kg}$

7. Calculate the boiling point elevation ($\Delta T_b$):

   • Using $K_b = 0.51 \text{ }^\circ\text{C kg/mol}$ for water: $\Delta T_b = 0.51 \text{ }^\circ\text{C kg/mol} \times 4.489 \text{ mol/kg} \approx 2.289 \text{ }^\circ\text{C}$

8. Calculate the normal boiling point of the solution:

   • $T_b = 100.0 \text{ }^\circ\text{C} + 2.289 \text{ }^\circ\text{C} \approx 102.289 \text{ }^\circ\text{C}$

Rounding to one decimal place, the normal boiling point is approximately $102.3 \text{ }^\circ\text{C}$.