A non conducting vessel thermally insulated from its surroundings, contains 100 g of water `0^(@)C`. The vessel is connected to a vacuum pump to pump to water vapour. As a result of this, some water is frozen. If the removal of water vapour is continued, what is the maximum amount of water that can be frozen in this manner? Laten heat of vaporisation of water `=22.5xx10^(5)Jkg^(-1)` and latent heat of fusion of ice `=3.36xx10^(5)Jkg^(-1)`

Answer

To solve the problem, we need to determine the maximum amount of water that can be frozen when water vapor is removed from a thermally insulated vessel containing 100 g of water at 0°C. We will use the principles of latent heat for this calculation.

### Step-by-Step Solution:

1. **Understand the Problem**: We have 100 g of water at 0°C in a thermally insulated vessel. When water vapor is removed, some of the water will freeze. We need to find out how much water can freeze.

2. **Identify Given Data**:
- Mass of water, \( m = 100 \, \text{g} = 0.1 \, \text{kg} \)
- Latent heat of fusion of ice, \( L_f = 3.36 \times 10^5 \, \text{J/kg} \)
- Latent heat of vaporization of water, \( L_v = 22.5 \times 10^5 \, \text{J/kg} \)

3. **Set Up the Energy Balance Equation**:
- Let \( x \) be the mass of water that freezes.
- The heat required to freeze \( x \) kg of water is given by:
\[ Q_{\text{freeze}} = x \cdot L_f \]
- The heat lost when the remaining water (which is \( (0.1 - x) \) kg) evaporates is given by:
\[ Q_{\text{evaporate}} = (0.1 - x) \cdot L_v \]

4. **Apply the Principle of Conservation of Energy**:
- The heat lost by the water that evaporates must equal the heat gained by the water that freezes:
\[ x \cdot L_f = (0.1 - x) \cdot L_v \]

5. **Substitute the Values**:
\[ x \cdot (3.36 \times 10^5) = (0.1 - x) \cdot (22.5 \times 10^5) \]

6. **Rearranging the Equation**:
\[ 3.36 \times 10^5 x = 2.25 \times 10^5 - 22.5 \times 10^5 x \]
\[ 3.36 \times 10^5 x + 22.5 \times 10^5 x = 2.25 \times 10^5 \]
\[ (3.36 + 22.5) \times 10^5 x = 2.25 \times 10^5 \]
\[ 25.86 \times 10^5 x = 2.25 \times 10^5 \]

7. **Solve for \( x \)**:
\[ x = \frac{2.25 \times 10^5}{25.86 \times 10^5} \]
\[ x = \frac{2.25}{25.86} \approx 0.0872 \, \text{kg} = 87.2 \, \text{g} \]

### Conclusion:
The maximum amount of water that can be frozen is approximately **87.2 g**.