The partial pressure of ethane over a solution containing `6.56xx10^(-2) g` of ethane is 1 bar. If the solution contains `5.00xx10^(-2) g` of ethane, then the partial pressure of the ethane gas will be

Answer

To solve the problem, we will use Henry's Law, which states that the amount of gas dissolved in a liquid is directly proportional to the partial pressure of that gas above the liquid at a constant temperature.

### Step-by-Step Solution:

1. **Identify the given values:**
- Mass of ethane in the first solution (m1) = \(6.56 \times 10^{-2} \, \text{g}\)
- Partial pressure of ethane in the first solution (P1) = \(1 \, \text{bar}\)
- Mass of ethane in the second solution (m2) = \(5.00 \times 10^{-2} \, \text{g}\)

2. **Write down Henry's Law relationship:**
According to Henry's Law:
\[ \frac{m_1}{m_2} = \frac{P_2}{P_1} \]
Rearranging this gives:
\[ P_2 = P_1 \times \frac{m_2}{m_1} \]

3. **Substitute the known values into the equation:**
\[ P_2 = 1 \, \text{bar} \times \frac{5.00 \times 10^{-2} \, \text{g}}{6.56 \times 10^{-2} \, \text{g}} \]

4. **Calculate the ratio of masses:**
\[ \frac{m_2}{m_1} = \frac{5.00 \times 10^{-2}}{6.56 \times 10^{-2}} \approx 0.763 \]

5. **Calculate P2:**
\[ P_2 = 1 \, \text{bar} \times 0.763 \approx 0.763 \, \text{bar} \]

### Final Answer:
The partial pressure of ethane gas when the solution contains \(5.00 \times 10^{-2} \, \text{g}\) of ethane is approximately **0.76 bar**.