A solution of sucross (molar mass = 342 g `"mol"^(-1)`) has been prepared by dissolving 68.4 g of sucrose in one kg of water. `K_(f)` for water is `"1.86 K kg mol"^(-1)` and vapour pressure of water at 298 K is 0.024 atm.
The mass of sodium chloride that should be dissolved in the same amount of water to get the same freezing point will be

Options

A. 136.8 g
B. 32.2 g
C. 5.85 g
D. 11.60 g

Explanation

To solve the problem, we need to determine the mass of sodium chloride (NaCl) that should be dissolved in 1 kg of water to achieve the same freezing point depression as the sucrose solution.

### Step-by-Step Solution:

1. **Calculate the Moles of Sucrose:**
- Given mass of sucrose = 68.4 g
- Molar mass of sucrose = 342 g/mol
- Number of moles of sucrose = Given mass / Molar mass
\[ \text{Number of moles of sucrose} = \frac{68.4 \, \text{g}}{342 \, \text{g/mol}} = 0.2 \, \text{mol} \]

2. **Calculate the Molality of the Sucrose Solution:**
- Molality (m) = Number of moles of solute / Mass of solvent (in kg)
- Mass of solvent (water) = 1 kg
\[ \text{Molality of sucrose} = \frac{0.2 \, \text{mol}}{1 \, \text{kg}} = 0.2 \, \text{mol/kg} \]

3. **Calculate the Freezing Point Depression (ΔTf) for Sucrose:**
- Freezing point depression formula:
\[ \Delta T_f = K_f \times m \]
- Where \( K_f \) for water = 1.86 K kg/mol
\[ \Delta T_f = 1.86 \, \text{K kg/mol} \times 0.2 \, \text{mol/kg} = 0.372 \, \text{K} \]

4. **Set Up the Equation for Sodium Chloride:**
- For sodium chloride, which dissociates into 2 ions (Na⁺ and Cl⁻), the van 't Hoff factor (i) = 2.
- The freezing point depression for NaCl will also be equal to ΔTf calculated for sucrose:
\[ \Delta T_f = i \cdot K_f \cdot m_{NaCl} \]
\[ 0.372 = 2 \cdot 1.86 \cdot m_{NaCl} \]

5. **Solve for the Molality of Sodium Chloride:**
\[ 0.372 = 3.72 \cdot m_{NaCl} \]
\[ m_{NaCl} = \frac{0.372}{3.72} = 0.1 \, \text{mol/kg} \]

6. **Calculate the Number of Moles of Sodium Chloride:**
- Since the mass of solvent is 1 kg, the number of moles of NaCl needed is:
\[ \text{Number of moles of NaCl} = 0.1 \, \text{mol} \]

7. **Calculate the Mass of Sodium Chloride:**
- Molar mass of NaCl = 58.5 g/mol
\[ \text{Mass of NaCl} = \text{Number of moles} \times \text{Molar mass} \]
\[ \text{Mass of NaCl} = 0.1 \, \text{mol} \times 58.5 \, \text{g/mol} = 5.85 \, \text{g} \]

### Final Answer:
The mass of sodium chloride that should be dissolved in 1 kg of water to achieve the same freezing point depression as the sucrose solution is **5.85 grams**.