Molal elevation constant and molal depression constant of water respectively (in K `m^(-1)`) are

Options

A. 0.52, 1.86
B. 1.86, 0.52
C. 1.52, 0.86
D. 0.86, 1.52

Answer

To solve the question regarding the molal elevation constant (Kb) and the molal depression constant (Kf) of water, we will follow these steps:

### Step 1: Understand the Definitions
- **Molal Elevation Constant (Kb)**: This is used to calculate the elevation in boiling point of a solvent when a solute is added.
- **Molal Depression Constant (Kf)**: This is used to calculate the depression in freezing point of a solvent when a solute is added.

### Step 2: Find Kb (Molal Elevation Constant)
1. **Formula for Kb**:
\[ Kb = \frac{R \cdot T_b^2}{1000 \cdot L_v} \]
where:
- \( R \) = Universal gas constant
- \( T_b \) = Boiling point of the solvent in Kelvin
- \( L_v \) = Latent heat of vaporization of the solvent

2. **Values for Water**:
- \( R = 2 \, \text{cal/K/mol} \)
- \( T_b = 100^\circ C = 373 \, K \) (convert Celsius to Kelvin by adding 273)
- \( L_v = 540 \, \text{kcal/kg} = 540 \times 1000 \, \text{cal/kg} \) (since 1 kcal = 1000 cal)

3. **Substituting Values**:
\[ Kb = \frac{2 \cdot (373)^2}{1000 \cdot 540 \times 1000} \]
\[ Kb = \frac{2 \cdot 139129}{540000000} \]
\[ Kb \approx 0.52 \, \text{K kg}^{-1} \text{mol}^{-1} \]

### Step 3: Find Kf (Molal Depression Constant)
1. **Formula for Kf**:
\[ Kf = \frac{R \cdot T_f^2}{1000 \cdot L_f} \]
where:
- \( T_f \) = Freezing point of the solvent in Kelvin
- \( L_f \) = Latent heat of fusion of the solvent

2. **Values for Water**:
- \( R = 2 \, \text{cal/K/mol} \)
- \( T_f = 0^\circ C = 273 \, K \)
- \( L_f = 80 \, \text{cal/g} = 80 \times 1000 \, \text{cal/kg} \)

3. **Substituting Values**:
\[ Kf = \frac{2 \cdot (273)^2}{1000 \cdot 80 \times 1000} \]
\[ Kf = \frac{2 \cdot 74529}{80000000} \]
\[ Kf \approx 1.86 \, \text{K kg}^{-1} \text{mol}^{-1} \]

### Final Answer
The molal elevation constant (Kb) and molal depression constant (Kf) of water are approximately:
- **Kb = 0.52 K kg\(^{-1}\) mol\(^{-1}\)**
- **Kf = 1.86 K kg\(^{-1}\) mol\(^{-1}\)**