An industrial waste water I found to contain 8.2% `Na_(3)PO_(4)` and 12% `MgSO_(4)` by mass in solution. If % ionisation of `Na_(3)PO_(4) "and"MgSO_(4)`Are 50 and 60 respectively then its normal boiliung point is [`K_(b)(H_(2)O) = 0.50 K kg "mol"^(-1)`] :

Options

A. `102.3^(@)C`
B. `103.35^(@)C`
C. `101.785^(@)`
D. none of these

Answer

To solve the problem of finding the normal boiling point of an industrial wastewater containing 8.2% Na₃PO₄ and 12% MgSO₄, we will follow these steps:

### Step 1: Calculate the Van't Hoff factors (i) for Na₃PO₄ and MgSO₄

1. **For Na₃PO₄:**
- Na₃PO₄ dissociates into 3 Na⁺ ions and 1 PO₄³⁻ ion.
- Therefore, the total number of ions produced = 3 + 1 = 4.
- Given that the percent ionization is 50%, we can express the Van't Hoff factor (i₁) as:
\[ i_1 = 1 + 3 \times 0.5 = 2.5 \]

2. **For MgSO₄:**
- MgSO₄ dissociates into 1 Mg²⁺ ion and 1 SO₄²⁻ ion.
- Therefore, the total number of ions produced = 1 + 1 = 2.
- Given that the percent ionization is 60%, we can express the Van't Hoff factor (i₂) as:
\[ i_2 = 1 + 0.6 = 1.6 \]

### Step 2: Calculate the molalities (m) of Na₃PO₄ and MgSO₄

1. **For Na₃PO₄:**
- The mass of Na₃PO₄ in 100 g of solution = 8.2 g.
- The mass of the solvent (water) = 100 g - 8.2 g = 91.8 g = 0.0918 kg.
- The molar mass of Na₃PO₄ = 164 g/mol.
- The number of moles of Na₃PO₄:
\[ \text{Moles of Na}_3\text{PO}_4 = \frac{8.2 \text{ g}}{164 \text{ g/mol}} = 0.0500 \text{ mol} \]
- The molality (m₁):
\[ m_1 = \frac{0.0500 \text{ mol}}{0.0918 \text{ kg}} = 0.544 \text{ mol/kg} \]

2. **For MgSO₄:**
- The mass of MgSO₄ in 100 g of solution = 12 g.
- The mass of the solvent (water) = 100 g - 12 g = 88 g = 0.088 kg.
- The molar mass of MgSO₄ = 120 g/mol.
- The number of moles of MgSO₄:
\[ \text{Moles of MgSO}_4 = \frac{12 \text{ g}}{120 \text{ g/mol}} = 0.100 \text{ mol} \]
- The molality (m₂):
\[ m_2 = \frac{0.100 \text{ mol}}{0.088 \text{ kg}} = 1.136 \text{ mol/kg} \]

### Step 3: Calculate the elevation in boiling point (ΔT_b)

Using the formula for elevation in boiling point:
\[ \Delta T_b = K_b \times (m_1 \cdot i_1 + m_2 \cdot i_2) \]
Where \( K_b = 0.50 \, \text{K kg/mol} \).

Substituting the values:
\[ \Delta T_b = 0.50 \times (0.544 \cdot 2.5 + 1.136 \cdot 1.6) \]
Calculating:
\[ \Delta T_b = 0.50 \times (1.36 + 1.818) = 0.50 \times 3.178 = 1.589 \, \text{K} \]

### Step 4: Calculate the normal boiling point of the solution

The normal boiling point of pure water is 100 °C. Therefore, the boiling point of the solution is:
\[ T_b = 100 + \Delta T_b = 100 + 1.589 = 101.589 \, \text{°C} \]

### Conclusion

The normal boiling point of the solution is approximately **101.59 °C**.

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