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

Options

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

Answer

To solve the problem, we will use Henry's Law, which states that the solubility of a gas in a liquid at a constant temperature is directly proportional to the partial pressure of that gas above the liquid. The formula can be expressed as:

\[ \frac{P_1}{P_2} = \frac{S_1}{S_2} \]

Where:
- \( P_1 \) = initial partial pressure of the gas
- \( P_2 \) = final partial pressure of the gas
- \( S_1 \) = solubility of the gas at \( P_1 \)
- \( S_2 \) = solubility of the gas at \( P_2 \)

### Step 1: Identify the known values
From the problem, we have:
- \( S_1 = 0.01 \, g/L \) (solubility at \( P_1 = 500 \, torr \))
- \( P_1 = 500 \, torr \)
- \( P_2 = 750 \, torr \)

### Step 2: Set up the equation using Henry's Law
Using the values we have:
\[ \frac{500 \, torr}{750 \, torr} = \frac{0.01 \, g/L}{S_2} \]

### Step 3: Cross-multiply to solve for \( S_2 \)
Cross-multiplying gives:
\[ 500 \, torr \cdot S_2 = 750 \, torr \cdot 0.01 \, g/L \]

### Step 4: Calculate the right side of the equation
Calculating the right side:
\[ 750 \, torr \cdot 0.01 \, g/L = 7.5 \, torr \cdot g/L \]

### Step 5: Isolate \( S_2 \)
Now, we can isolate \( S_2 \):
\[ S_2 = \frac{7.5 \, torr \cdot g/L}{500 \, torr} \]

### Step 6: Perform the division
Calculating \( S_2 \):
\[ S_2 = \frac{7.5}{500} \, g/L = 0.015 \, g/L \]

### Conclusion
Thus, the solubility of nitrogen in water at 750 torr partial pressure is \( 0.015 \, g/L \).

### Final Answer
The solubility at 750 torr is **0.015 g/L**.

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