The solubility of `N_(2)` in water at 300K at 300K and 500 torr partial pressure `0.01 gL^(-1)`. The solubility (in `g L^(-1)` ) at 750 torr partial pressure is `:`

Options

A. 0.0075
B. 0.005
C. 0.02
D. 0.015

Answer

To solve the problem of determining the solubility of nitrogen gas (N₂) in water at a partial pressure of 750 torr, 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 500 torr (S1) = 0.01 g/L
- Partial pressure 1 (P1) = 500 torr
- Partial pressure 2 (P2) = 750 torr
- We need to find the solubility at 750 torr (S2).

2. **Apply Henry's Law:**
According to Henry's law:
\[ \frac{P1}{P2} = \frac{S1}{S2} \]
Rearranging this gives us:
\[ S2 = S1 \times \frac{P2}{P1} \]

3. **Substitute the Known Values:**
Now, we substitute the known values into the equation:
\[ S2 = 0.01 \, \text{g/L} \times \frac{750 \, \text{torr}}{500 \, \text{torr}} \]

4. **Calculate S2:**
\[ S2 = 0.01 \, \text{g/L} \times 1.5 = 0.015 \, \text{g/L} \]

5. **Final Result:**
The solubility of nitrogen gas at a partial pressure of 750 torr is:
\[ S2 = 0.015 \, \text{g/L} \]

### Answer:
The solubility at 750 torr partial pressure is **0.015 g/L**.