A water sample contains 9.5% `MgCl_(2)` and 11.7% NaCl (by weight).Assuming 80% ionisation of each salt boiling point of water will be `(K_(b)=0.52)`

Options

A. `110.01^(@)C`
B. `376.22 K`
C. 277.25 K
D. `102.5^(@)C`

Answer

To solve the problem, we need to calculate the boiling point elevation of a water sample containing 9.5% MgCl₂ and 11.7% NaCl by weight, considering 80% ionization of each salt and a given Kb value of 0.52.

### Step-by-Step Solution:

1. **Calculate the Mass of Solvent:**
- Total mass of the solution = 100 g
- Mass of MgCl₂ = 9.5 g
- Mass of NaCl = 11.7 g
- Mass of solvent (water) = Total mass - (Mass of MgCl₂ + Mass of NaCl)
\[ \text{Mass of solvent} = 100 - (9.5 + 11.7) = 100 - 21.2 = 78.8 \text{ g} \]

2. **Calculate Moles of Each Salt:**
- **For MgCl₂:**
- Molar mass of MgCl₂ = 24 (Mg) + 2 × 35.5 (Cl) = 95 g/mol
- Moles of MgCl₂ = \(\frac{9.5 \text{ g}}{95 \text{ g/mol}} = 0.1 \text{ moles}\)

- **For NaCl:**
- Molar mass of NaCl = 23 (Na) + 35.5 (Cl) = 58.5 g/mol
- Moles of NaCl = \(\frac{11.7 \text{ g}}{58.5 \text{ g/mol}} = 0.2 \text{ moles}\)

3. **Calculate the Van't Hoff Factor (i):**
- **For NaCl:**
- NaCl dissociates into Na⁺ and Cl⁻.
- Degree of ionization (α) = 0.8
- Van't Hoff factor (i) for NaCl = \(1 + \alpha = 1 + 0.8 = 1.8\)

- **For MgCl₂:**
- MgCl₂ dissociates into Mg²⁺ and 2 Cl⁻.
- Van't Hoff factor (i) for MgCl₂ = \(1 + 2\alpha = 1 + 2(0.8) = 2.6\)

4. **Calculate Molality (M):**
- Molality (m) is defined as moles of solute per kg of solvent.
- Mass of solvent = 78.8 g = 0.0788 kg
- Molality of NaCl = \(\frac{0.2 \text{ moles}}{0.0788 \text{ kg}} = 2.54 \text{ mol/kg}\)
- Molality of MgCl₂ = \(\frac{0.1 \text{ moles}}{0.0788 \text{ kg}} = 1.27 \text{ mol/kg}\)

5. **Calculate the Boiling Point Elevation (ΔTb):**
- Using the formula:
\[ \Delta T_b = i \cdot K_b \cdot m \]
- For NaCl:
\[ \Delta T_{b, \text{NaCl}} = 1.8 \cdot 0.52 \cdot 2.54 = 2.36 \text{ °C} \]
- For MgCl₂:
\[ \Delta T_{b, \text{MgCl₂}} = 2.6 \cdot 0.52 \cdot 1.27 = 1.73 \text{ °C} \]

6. **Total Boiling Point Elevation:**
\[ \Delta T_b = \Delta T_{b, \text{NaCl}} + \Delta T_{b, \text{MgCl₂}} = 2.36 + 1.73 = 4.09 \text{ °C} \]

7. **Calculate the New Boiling Point:**
- Normal boiling point of water = 100 °C
- New boiling point = 100 °C + ΔTb
\[ \text{New boiling point} = 100 + 4.09 = 104.09 \text{ °C} \]

### Final Answer:
The boiling point of the water sample is approximately **104.09 °C**.