In `10^3` Litre sample of hard water `CaSO_4 and MgSO_4` is present.If elevation in Boiling point is `0.000052^@C`.Calculate the degree of Hardness of hard water .(`K_b` for `H_2O=0.52`)

Explanation

To calculate the degree of hardness of hard water containing CaSO₄ and MgSO₄ based on the elevation in boiling point, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the given data:**
- Volume of hard water sample = \(10^3\) L
- Elevation in boiling point (\( \Delta T_b \)) = \(0.000052^\circ C\)
- \(K_b\) for water = \(0.52 \, \text{°C kg/mol}\)

2. **Identify the van't Hoff factors (\(i\)):**
- For CaSO₄, it dissociates into 2 ions: \(Ca^{2+}\) and \(SO_4^{2-}\). Thus, \(i_{CaSO_4} = 2\).
- For MgSO₄, it also dissociates into 2 ions: \(Mg^{2+}\) and \(SO_4^{2-}\). Thus, \(i_{MgSO_4} = 2\).

3. **Set up the equation for elevation in boiling point:**
\[ \Delta T_b = i_1 \cdot K_b \cdot m_1 + i_2 \cdot K_b \cdot m_2 \]
Here, \(m_1\) and \(m_2\) are the molalities of CaSO₄ and MgSO₄ respectively.

4. **Substituting the values:**
\[ 0.000052 = 2 \cdot 0.52 \cdot m_1 + 2 \cdot 0.52 \cdot m_2 \]
This simplifies to:
\[ 0.000052 = 2 \cdot 0.52 (m_1 + m_2) \]

5. **Calculate \(m_1 + m_2\):**
\[ 0.000052 = 1.04 (m_1 + m_2) \]
\[ m_1 + m_2 = \frac{0.000052}{1.04} = 0.00005 \, \text{mol/kg} \]

6. **Convert the volume of water to kg:**
- Since the density of water is approximately \(1 \, \text{kg/L}\), \(10^3 \, \text{L}\) of water is \(10^3 \, \text{kg}\).

7. **Calculate the total moles of solute:**
\[ n_1 + n_2 = (m_1 + m_2) \cdot \text{mass of solvent} = 0.00005 \cdot 10^3 = 0.05 \, \text{mol} \]

8. **Convert moles to grams of CaCO₃ equivalent:**
- The equivalent weight of CaCO₃ is \(100 \, \text{g/mol}\).
\[ \text{grams of CaCO₃} = 0.05 \, \text{mol} \times 100 \, \text{g/mol} = 5 \, \text{g} \]

9. **Calculate the degree of hardness in parts per million (ppm):**
\[ \text{Degree of hardness} = \frac{\text{grams of CaCO₃}}{\text{volume of water in L}} \times 10^6 = \frac{5 \, \text{g}}{10^6 \, \text{g}} \times 10^6 = 5 \, \text{ppm} \]

### Final Answer:
The degree of hardness of hard water is **5 parts per million (ppm)**.