The solubility of `N_(2)(g)`in water exposed to the atmosphere, when the partial pressure is `593`mm is `5.3xx10^(-4)`M. Its solubility at 760 mm and at the same temperature is

Options

A. `4.1xx10^(-4)M`
B. `6.8xx10^(-4)M`
C. `1500M`
D. `2400M`

Explanation

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

### Step-by-Step Solution:

1. **Identify Given Values:**
- Solubility at partial pressure \( P_1 = 593 \, \text{mmHg} \) is \( S_1 = 5.3 \times 10^{-4} \, \text{M} \).
- We need to find the solubility \( S_2 \) at \( P_2 = 760 \, \text{mmHg} \).

2. **Apply Henry's Law:**
According to Henry's Law:
\[ \frac{S_1}{S_2} = \frac{P_1}{P_2} \]

3. **Substitute the Known Values:**
Substitute the known values into the equation:
\[ \frac{5.3 \times 10^{-4}}{S_2} = \frac{593}{760} \]

4. **Cross-Multiply to Solve for \( S_2 \):**
Rearranging gives:
\[ S_2 = \frac{5.3 \times 10^{-4} \times 760}{593} \]

5. **Calculate \( S_2 \):**
Now, perform the calculation:
\[ S_2 = \frac{5.3 \times 10^{-4} \times 760}{593} \approx 6.8 \times 10^{-4} \, \text{M} \]

6. **Conclusion:**
The solubility of \( N_2(g) \) in water at a partial pressure of \( 760 \, \text{mmHg} \) is approximately \( 6.8 \times 10^{-4} \, \text{M} \).

### Final Answer:
The solubility of \( N_2(g) \) at \( 760 \, \text{mmHg} \) is \( 6.8 \times 10^{-4} \, \text{M} \).