A `500gm` water consist of `15gm` ethane at any temp. `T`, at a pressure `=2` atm. Find Pressure of gas required to dissolve `30gm` gas in `300gm` water.

Answer

To solve the problem, we will use Henry's Law, which states that the partial pressure of a gas in a solution is proportional to the mole fraction of that gas in the solution.

### Step-by-Step Solution:

1. **Calculate the number of moles of ethane in the first case:**
- Given mass of ethane (C2H6) = 15 g
- Molar mass of ethane = 30 g/mol
- Number of moles of ethane = mass / molar mass = 15 g / 30 g/mol = 0.5 mol

2. **Calculate the number of moles of water in the first case:**
- Given mass of water = 500 g - 15 g = 485 g
- Molar mass of water = 18 g/mol
- Number of moles of water = 485 g / 18 g/mol ≈ 26.94 mol

3. **Calculate the mole fraction of ethane in the first case:**
- Mole fraction of ethane (X_ethane) = moles of ethane / (moles of ethane + moles of water)
- X_ethane = 0.5 / (0.5 + 26.94) ≈ 0.018

4. **Use Henry's Law to find the partial pressure of ethane in the first case:**
- Given pressure (P1) = 2 atm
- P1 = K_H * X_ethane
- Therefore, K_H = P1 / X_ethane = 2 atm / 0.018 ≈ 111.11 atm

5. **Calculate the number of moles of ethane in the second case:**
- Given mass of ethane = 30 g
- Number of moles of ethane = 30 g / 30 g/mol = 1 mol

6. **Calculate the number of moles of water in the second case:**
- Given mass of water = 300 g
- Number of moles of water = 300 g / 18 g/mol ≈ 16.67 mol

7. **Calculate the mole fraction of ethane in the second case:**
- X_ethane (second case) = moles of ethane / (moles of ethane + moles of water)
- X_ethane = 1 / (1 + 16.67) ≈ 0.057

8. **Use Henry's Law to find the partial pressure of ethane in the second case:**
- P2 = K_H * X_ethane (second case)
- P2 = 111.11 atm * 0.057 ≈ 6.33 atm

### Final Answer:
The pressure of the gas required to dissolve 30 g of ethane in 300 g of water is approximately **6.33 atm**.

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