At `300 K, 40 mL` of `O_(3)(g)` dissolves in `100g` of water at `1.0atm`. What mass of ozone dissolved in `400g ` of water at a pressure of `4.0atm` at `300 K` ?

Options

A. `0.1 g`
B. `1.2g`
C. `0.48g`
D. `4.8g`

Explanation

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

\[ C \propto P \]
or
\[ C = k \cdot P \]

Where:
- \( C \) is the concentration of the gas,
- \( P \) is the pressure of the gas,
- \( k \) is a constant.

### Step 1: Determine the concentration of ozone in the first case.

Given:
- Volume of ozone (\( V_1 \)) = 40 mL
- Mass of water (\( m_1 \)) = 100 g
- Pressure (\( P_1 \)) = 1.0 atm

To find the concentration \( C_1 \) of ozone in the first case, we can use the formula:

\[ C_1 = \frac{V_1}{m_1} \]

Substituting the values:

\[ C_1 = \frac{40 \text{ mL}}{100 \text{ g}} = 0.4 \text{ mL/g} \]

### Step 2: Set up the ratio using Henry's Law.

Using Henry's Law, we can set up the following ratio:

\[ \frac{C_1}{P_1} = \frac{C_2}{P_2} \]

Where:
- \( C_2 \) is the concentration of ozone in the second case,
- \( P_2 \) = 4.0 atm.

### Step 3: Express \( C_2 \) in terms of \( X \).

Let \( X \) be the volume of ozone that dissolves in 400 g of water. The concentration \( C_2 \) can be expressed as:

\[ C_2 = \frac{X}{400 \text{ g}} \]

### Step 4: Substitute into the ratio.

Now substituting into the ratio:

\[ \frac{0.4 \text{ mL/g}}{1.0 \text{ atm}} = \frac{X/400 \text{ g}}{4.0 \text{ atm}} \]

### Step 5: Cross-multiply and solve for \( X \).

Cross-multiplying gives:

\[ 0.4 \cdot 4.0 = X \cdot \frac{1.0}{400} \]

This simplifies to:

\[ 1.6 = \frac{X}{400} \]

Now, multiplying both sides by 400:

\[ X = 1.6 \cdot 400 = 640 \text{ mL} \]

### Step 6: Convert volume to moles.

Now we need to convert the volume of ozone to moles using the ideal gas law:

\[ PV = nRT \]

Rearranging gives:

\[ n = \frac{PV}{RT} \]

Given:
- \( P = 4 \text{ atm} \)
- \( V = 640 \text{ mL} = 0.640 \text{ L} \)
- \( R = 0.0821 \text{ L atm/(K mol)} \)
- \( T = 300 \text{ K} \)

Substituting the values:

\[ n = \frac{4 \cdot 0.640}{0.0821 \cdot 300} \]

Calculating gives:

\[ n = \frac{2.56}{24.63} \approx 0.104 \text{ moles} \]

### Step 7: Calculate the mass of ozone.

The molar mass of ozone (\( O_3 \)) is:

\[ 3 \times 16 \text{ g/mol} = 48 \text{ g/mol} \]

Now, the mass of ozone can be calculated as:

\[ \text{Mass} = n \cdot \text{Molar Mass} = 0.104 \cdot 48 \text{ g} \approx 4.992 \text{ g} \]

### Conclusion

Thus, the mass of ozone that can be dissolved in 400 g of water at a pressure of 4.0 atm at 300 K is approximately **5.0 g**.

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