The mole fraction of a gas dissolved in a solvent is given by Henry's law. Constant for gas in water at 298 K is `5.55xx10^(7)` Torr and the partial pressure of the gas is 200 Torr, then what is the amount of the gas dissolved in 1.0 kg of water?

Options

A. `2.0xx10^(-4)mol`
B. `2.5xx10^(-5)mol`
C. `3.7xx10^(-6)mol`
D. `1.2xx10^(-8)mol`

Explanation

To solve the problem, we will use Henry's law and the concept of mole fraction to find the amount of gas dissolved in 1.0 kg of water.

### Step-by-Step Solution:

1. **Understand Henry's Law**:
Henry's law states that the amount of gas dissolved in a liquid is proportional to the partial pressure of that gas above the liquid. The relationship is given by:
\[ P = k_H \cdot x \]
where \( P \) is the partial pressure of the gas, \( k_H \) is Henry's law constant, and \( x \) is the mole fraction of the gas in the solvent.

2. **Rearranging Henry's Law**:
We can rearrange the equation to find the mole fraction \( x \):
\[ x = \frac{P}{k_H} \]

3. **Substituting the Given Values**:
We know:
- Partial pressure \( P = 200 \) Torr
- Henry's law constant \( k_H = 5.55 \times 10^7 \) Torr

Substituting these values into the equation:
\[ x = \frac{200}{5.55 \times 10^7} \]

4. **Calculating the Mole Fraction**:
Performing the calculation:
\[ x = \frac{200}{5.55 \times 10^7} \approx 3.6 \times 10^{-6} \]

5. **Understanding Mole Fraction**:
The mole fraction \( x \) is defined as:
\[ x = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{solvent}}} \]
Since the number of moles of solute (gas) is very small compared to the number of moles of solvent (water), we can approximate:
\[ x \approx \frac{n_{\text{solute}}}{n_{\text{solvent}}} \]

6. **Calculating Moles of Solvent**:
The number of moles of water can be calculated using the formula:
\[ n_{\text{solvent}} = \frac{\text{mass of solvent}}{\text{molar mass of solvent}} \]
Given that the mass of water is 1.0 kg (or 1000 g) and the molar mass of water is 18 g/mol:
\[ n_{\text{solvent}} = \frac{1000 \text{ g}}{18 \text{ g/mol}} \approx 55.56 \text{ moles} \]

7. **Finding Moles of Solute**:
Now we can use the mole fraction to find the number of moles of the gas (solute):
\[ x = \frac{n_{\text{solute}}}{n_{\text{solvent}}} \]
Rearranging gives:
\[ n_{\text{solute}} = x \cdot n_{\text{solvent}} \]
Substituting the values:
\[ n_{\text{solute}} = 3.6 \times 10^{-6} \times 55.56 \approx 2 \times 10^{-4} \text{ moles} \]

8. **Final Result**:
The amount of gas dissolved in 1.0 kg of water is approximately:
\[ n_{\text{solute}} \approx 2 \times 10^{-4} \text{ moles} \]