The solubility of a gas in water at 300 K under a pressure of 100 atmospheres is `4xx10^(-3)"kg L"^(-1)`. Therefore, the mass of the gas in kg dissolved in 250 mL of water under a pressure of 250 atmospheres at 300 K is

Options

A. `2.5xx10^(-3)`
B. `2.0xx10^(-3)`
C. `1.25xx10^(-3)`
D. `5.0xx10^(-3)`

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 pressure of the gas above the liquid.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Solubility of gas at 100 atm: \( M_1 = 4 \times 10^{-3} \, \text{kg L}^{-1} \)
- Pressure for the first case: \( P_1 = 100 \, \text{atm} \)
- Pressure for the second case: \( P_2 = 250 \, \text{atm} \)
- Volume of water: \( V = 250 \, \text{mL} = 0.250 \, \text{L} \)

2. **Apply Henry's Law:**
According to Henry's law:
\[ \frac{M_1}{M_2} = \frac{P_1}{P_2} \]
where \( M_2 \) is the solubility of the gas at the second pressure.

3. **Rearrange the Equation:**
We can rearrange the equation to solve for \( M_2 \):
\[ M_2 = M_1 \times \frac{P_2}{P_1} \]

4. **Substitute the Known Values:**
Substitute \( M_1 \), \( P_1 \), and \( P_2 \) into the equation:
\[ M_2 = 4 \times 10^{-3} \, \text{kg L}^{-1} \times \frac{250 \, \text{atm}}{100 \, \text{atm}} \]

5. **Calculate \( M_2 \):**
\[ M_2 = 4 \times 10^{-3} \, \text{kg L}^{-1} \times 2.5 = 10 \times 10^{-3} \, \text{kg L}^{-1} = 1 \times 10^{-2} \, \text{kg L}^{-1} \]

6. **Calculate the Mass of Gas in 250 mL of Water:**
Since \( M_2 \) is the solubility in kg per liter, we can find the mass of gas dissolved in 0.250 L:
\[ \text{Mass} = M_2 \times V = 1 \times 10^{-2} \, \text{kg L}^{-1} \times 0.250 \, \text{L} = 2.5 \times 10^{-3} \, \text{kg} \]

### Final Answer:
The mass of the gas dissolved in 250 mL of water under a pressure of 250 atmospheres at 300 K is \( 2.5 \times 10^{-3} \, \text{kg} \).