The partial pressure of ethane over a solution containing `6.56xx10^(-3) "g"` of ethane is 1 bar. If the solution contains `5xx10^(-2)g` of ethane, then what will be the partial pressure of gas ?

Options

A. 4.2 bar
B. 7.6 bar
C. 8.9 bar
D. 9.8 bar

Explanation

To solve this problem, we will use Henry's law, which states that the mass of a gas dissolved in a liquid is directly proportional to the partial pressure of the gas above the liquid. Let's go through the steps to find the partial pressure of ethane when the mass of ethane is increased.

### Step-by-Step Solution:

1. **Identify the given values:**
- Initial mass of ethane, \( m_1 = 6.56 \times 10^{-3} \) grams
- Initial partial pressure of ethane, \( P_1 = 1 \) bar
- New mass of ethane, \( m_2 = 5 \times 10^{-2} \) grams

2. **Apply Henry's law:**
According to Henry's law:
\[ \frac{m_1}{P_1} = \frac{m_2}{P_2} \]
Here, \( P_2 \) is the new partial pressure we need to find.

3. **Rearrange the equation to solve for \( P_2 \):**
\[ P_2 = P_1 \times \frac{m_2}{m_1} \]

4. **Substitute the known values into the equation:**
\[ P_2 = 1 \, \text{bar} \times \frac{5 \times 10^{-2} \, \text{g}}{6.56 \times 10^{-3} \, \text{g}} \]

5. **Perform the division to find the ratio:**
\[ \frac{5 \times 10^{-2}}{6.56 \times 10^{-3}} = \frac{5}{6.56} \times 10^{1} = 0.762 \times 10^{1} = 7.62 \]

6. **Calculate the new partial pressure:**
\[ P_2 = 1 \, \text{bar} \times 7.62 = 7.62 \, \text{bar} \]

### Final Answer:
The partial pressure of ethane when the solution contains \( 5 \times 10^{-2} \) grams of ethane is \( 7.62 \) bar.