A thermally isolated vessel contains `100`g of water at `0^(@)C` when air above the water is pumped out, some of the water freezes and some evaporates at `0^(@)C` itself. Calculate the mass at `0^(@)C=2.10xx10^(6) j//kg` and latent heat of fusion of ice `=3.36xx10^(5) j//kg`.

Explanation

To solve the problem, we need to establish a balance between the heat gained by the water that evaporates and the heat lost by the water that freezes.

### Step-by-Step Solution:

1. **Define Variables**:
- Let \( m \) be the mass of water that freezes (in grams).
- The mass of water that evaporates will then be \( 100 - m \) grams.

2. **Convert Mass to Kilograms**:
- Since the specific heat and latent heat are given in joules per kilogram, we need to convert grams to kilograms:
\[ m \text{ (in kg)} = \frac{m}{1000} \]
\[ (100 - m) \text{ (in kg)} = \frac{100 - m}{1000} \]

3. **Calculate Heat Lost by Freezing Water**:
- The heat lost by the water that freezes can be calculated using the latent heat of fusion:
\[ Q_{\text{freeze}} = m \times L_f = m \times 3.36 \times 10^5 \text{ J/kg} \]

4. **Calculate Heat Gained by Evaporating Water**:
- The heat gained by the water that evaporates can be calculated using the specific heat of water:
\[ Q_{\text{evaporate}} = (100 - m) \times L_e = (100 - m) \times 2.1 \times 10^6 \text{ J/kg} \]

5. **Set Up the Equation**:
- Since the heat lost by the freezing water equals the heat gained by the evaporating water, we can set up the equation:
\[ m \times 3.36 \times 10^5 = (100 - m) \times 2.1 \times 10^6 \]

6. **Expand and Rearrange the Equation**:
- Expanding the right side:
\[ m \times 3.36 \times 10^5 = 2.1 \times 10^6 \times 100 - 2.1 \times 10^6 \times m \]
- Rearranging gives:
\[ m \times (3.36 \times 10^5 + 2.1 \times 10^6) = 2.1 \times 10^8 \]

7. **Solve for \( m \)**:
- Combine the terms:
\[ m \times (2.1 \times 10^6 + 3.36 \times 10^5) = 2.1 \times 10^8 \]
- Calculate \( 2.1 \times 10^6 + 3.36 \times 10^5 = 2.436 \times 10^6 \):
\[ m \times 2.436 \times 10^6 = 2.1 \times 10^8 \]
- Now, solve for \( m \):
\[ m = \frac{2.1 \times 10^8}{2.436 \times 10^6} \approx 86.2 \text{ grams} \]

### Final Result:
The mass of the ice formed is approximately \( 86.2 \) grams.