On a hill station water boils at `98^(@)C`. Amount of salt (NaCl) which should be addedto make its boilings point `100^(@)C` is (`K_(b)=0.52" K kg mole"`)

Options

A. 112.5 g/L
B. 281.2 g/L
C. 225 g/L
D. 140.6 g/L

Answer

To solve the problem, we need to determine the amount of salt (NaCl) that should be added to water to raise its boiling point from 98°C to 100°C. We will use the formula for boiling point elevation.

### Step-by-step Solution:

1. **Identify the given values:**
- Initial boiling point of water (Tb solvent) = 98°C
- Final boiling point of solution (Tb solution) = 100°C
- Boiling point elevation constant (Kb) = 0.52 K kg/mol
- Van't Hoff factor (i) for NaCl = 2 (because NaCl dissociates into Na⁺ and Cl⁻)

2. **Calculate the change in boiling point (ΔTb):**
\[ \Delta T_b = T_b \text{ solution} - T_b \text{ solvent} = 100°C - 98°C = 2°C \]

3. **Use the boiling point elevation formula:**
\[ \Delta T_b = i \cdot K_b \cdot m \]
where:
- \( m \) = molality of the solution (moles of solute per kg of solvent)

4. **Rearranging the formula to find molality (m):**
\[ m = \frac{\Delta T_b}{i \cdot K_b} \]
Substituting the known values:
\[ m = \frac{2°C}{2 \cdot 0.52 \, \text{K kg/mol}} = \frac{2}{1.04} \approx 1.923 \, \text{mol/kg} \]

5. **Calculate the number of moles of NaCl needed:**
Since molality is defined as moles of solute per kg of solvent, we need to find the number of moles for 1 kg of solvent:
\[ \text{Moles of NaCl} = m \cdot \text{mass of solvent (kg)} = 1.923 \, \text{mol/kg} \cdot 1 \, \text{kg} = 1.923 \, \text{mol} \]

6. **Calculate the mass of NaCl needed:**
The molar mass of NaCl is approximately 58.44 g/mol. Therefore, the mass of NaCl required is:
\[ \text{Mass of NaCl} = \text{moles} \cdot \text{molar mass} = 1.923 \, \text{mol} \cdot 58.44 \, \text{g/mol} \approx 112.5 \, \text{g} \]

7. **Conclusion:**
The amount of NaCl that should be added to the water to raise its boiling point to 100°C is approximately **112.5 grams**.