Sea water is found to contain `5.85% NaCI` and`9.50% MgCI_(2)` by weight of solution. Calculate its normal boiling point assuming `80%` ionisation for `NaCI` and `50%` ionisation of `MgCI[K_(b)(H_(2)O) =0.51 kg mol^(-1)K]`

Options

A. `T_(b)=101.9^(@)c`
B. `T_(b)=102.3^(@)C`
C. `T_(b)=108.5^(@)C`
D. `T_(b)=110.3^(@)C`

Explanation

To solve the problem of calculating the normal boiling point of seawater containing 5.85% NaCl and 9.50% MgCl₂, we will follow these steps:

### Step 1: Calculate the mass of solutes in 100 g of seawater
In 100 g of seawater:
- Mass of NaCl = 5.85 g
- Mass of MgCl₂ = 9.50 g

### Step 2: Calculate the number of moles of each solute
- **Molar mass of NaCl** = 58.5 g/mol
- **Molar mass of MgCl₂** = 95 g/mol

**Number of moles of NaCl:**
\[ \text{Moles of NaCl} = \frac{5.85 \text{ g}}{58.5 \text{ g/mol}} = 0.1 \text{ moles} \]

**Number of moles of MgCl₂:**
\[ \text{Moles of MgCl₂} = \frac{9.50 \text{ g}}{95 \text{ g/mol}} = 0.1 \text{ moles} \]

### Step 3: Calculate the dissociation of each solute
- **NaCl dissociates into Na⁺ and Cl⁻.**
- Given 80% ionization for NaCl:
- Moles of Na⁺ formed = \(0.1 \times 0.8 = 0.08\)
- Moles of Cl⁻ formed = \(0.1 \times 0.8 = 0.08\)
- Remaining NaCl = \(0.1 - 0.08 = 0.02\)

Total moles from NaCl after dissociation:
\[ \text{Total moles from NaCl} = 0.08 + 0.08 + 0.02 = 0.18 \]

- **MgCl₂ dissociates into Mg²⁺ and 2 Cl⁻.**
- Given 50% ionization for MgCl₂:
- Moles of Mg²⁺ formed = \(0.1 \times 0.5 = 0.05\)
- Moles of Cl⁻ formed = \(0.1 \times 0.5 \times 2 = 0.10\)
- Remaining MgCl₂ = \(0.1 - 0.05 = 0.05\)

Total moles from MgCl₂ after dissociation:
\[ \text{Total moles from MgCl₂} = 0.05 + 0.10 + 0.05 = 0.20 \]

### Step 4: Calculate the total number of moles in the solution
\[ \text{Total moles} = 0.18 + 0.20 = 0.38 \]

### Step 5: Calculate the mass of the solvent (water)
Total mass of seawater = 100 g
Mass of solutes = \(5.85 \text{ g} + 9.50 \text{ g} = 15.35 \text{ g}\)

Mass of solvent (water) = \(100 \text{ g} - 15.35 \text{ g} = 84.65 \text{ g}\)

### Step 6: Calculate the molality of the solution
\[ \text{Molality (m)} = \frac{\text{Number of moles of solute}}{\text{Mass of solvent in kg}} = \frac{0.38}{0.08465} \approx 4.49 \text{ mol/kg} \]

### Step 7: Calculate the elevation in boiling point
Using the formula:
\[ \Delta T_b = K_b \cdot m \]
Where \(K_b = 0.51 \text{ kg/mol/K}\):
\[ \Delta T_b = 0.51 \cdot 4.49 \approx 2.29 \text{ K} \]

### Step 8: Calculate the normal boiling point of seawater
Normal boiling point of water = 100 °C
\[ \text{Normal boiling point of seawater} = 100 + 2.29 \approx 102.3 \text{ °C} \]

### Final Answer
The normal boiling point of seawater is approximately **102.3 °C**.

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