The boiling point of water of 735 torr is `99.07^@C` The mass of NaCl added in 100g water to make its boiling point `100^@C` is

Options

A. 10.68 g
B. 5.34g
C. 2.67g
D. 26.7g

Answer

To solve the problem of determining the mass of NaCl needed to raise the boiling point of 100 g of water from 99.07 °C to 100 °C, we can follow these steps:

### Step-by-Step Solution:

1. **Determine the Elevation in Boiling Point (ΔT):**
\[ \Delta T = T_{\text{solution}} - T_{\text{pure solvent}} = 100 °C - 99.07 °C = 0.93 °C \]

2. **Use the Boiling Point Elevation Formula:**
The formula for boiling point elevation is given by:
\[ \Delta T = i \cdot K_b \cdot m \]
Where:
- \( \Delta T \) = elevation in boiling point
- \( i \) = van 't Hoff factor (number of particles the solute dissociates into)
- \( K_b \) = ebullioscopic constant of the solvent (for water, \( K_b = 0.52 \, °C \, kg/mol \))
- \( m \) = molality of the solution

3. **Determine the van 't Hoff Factor (i) for NaCl:**
NaCl dissociates into two ions: Na⁺ and Cl⁻. Therefore, the van 't Hoff factor \( i \) is:
\[ i = 2 \]

4. **Rearrange the Boiling Point Elevation Formula to Find Molality (m):**
\[ m = \frac{\Delta T}{i \cdot K_b} \]
Substituting the known values:
\[ m = \frac{0.93}{2 \cdot 0.52} = \frac{0.93}{1.04} \approx 0.893 mol/kg \]

5. **Convert Molality to Moles of NaCl:**
Since molality is defined as moles of solute per kg of solvent, we need to find the moles of NaCl in 100 g of water (0.1 kg):
\[ \text{Moles of NaCl} = m \cdot \text{mass of solvent (kg)} = 0.893 \cdot 0.1 \approx 0.0893 \, \text{moles} \]

6. **Calculate the Mass of NaCl Required:**
The molar mass of NaCl is approximately 58.5 g/mol. Therefore, the mass of NaCl needed is:
\[ \text{Mass of NaCl} = \text{moles} \cdot \text{molar mass} = 0.0893 \, \text{moles} \cdot 58.5 \, \text{g/mol} \approx 5.22 \, \text{g} \]

### Final Answer:
The mass of NaCl added to 100 g of water to raise its boiling point to 100 °C is approximately **5.22 g**.