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` ?
a. 0.1 g
b. 1.24 g
c. 0.48 g
d. 4.8 g

Options

A. 0.1 g
B. 1.24 g
C. 0.48 g
D. 4.8 g

Answer

To solve the problem, we will use Henry's law, which states that the concentration of a gas in a liquid is directly proportional to the pressure of that gas above the liquid.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- For the first case:
- Volume of ozone dissolved (V1) = 40 mL
- Mass of water (m1) = 100 g
- Pressure (P1) = 1.0 atm
- For the second case:
- Mass of water (m2) = 400 g
- Pressure (P2) = 4.0 atm
- We need to find the mass of ozone dissolved (x) in this case.

2. **Calculate the Concentration in the First Case:**
- Concentration (C1) can be calculated as:
\[ C1 = \frac{V1}{m1} = \frac{40 \text{ mL}}{100 \text{ g}} = 0.4 \text{ mL/g} \]

3. **Set Up the Ratio Using Henry's Law:**
- According to Henry's law:
\[ \frac{C1}{P1} = \frac{C2}{P2} \]
- Here, C2 is the concentration of ozone in the second case, which can be expressed as:
\[ C2 = \frac{x}{400 \text{ g}} \text{ (where x is the volume of ozone in mL)} \]

4. **Substitute the Known Values:**
- Plugging in the values we have:
\[ \frac{0.4 \text{ mL/g}}{1.0 \text{ atm}} = \frac{\frac{x}{400 \text{ g}}}{4.0 \text{ atm}} \]

5. **Cross Multiply to Solve for x:**
- Cross multiplying gives:
\[ 0.4 \times 4.0 = x \times 1.0 / 400 \]
- Simplifying this:
\[ 1.6 = \frac{x}{400} \]
- Therefore:
\[ x = 1.6 \times 400 = 640 \text{ mL} \]

6. **Convert Volume to Moles:**
- Use the ideal gas equation \( PV = nRT \) to find the number of moles (n):
\[ n = \frac{PV}{RT} \]
- Where:
- P = 4.0 atm
- V = 0.640 L (since 640 mL = 0.640 L)
- R = 0.0821 L·atm/(K·mol)
- T = 300 K
- Plugging in the values:
\[ n = \frac{(4.0 \text{ atm}) \times (0.640 \text{ L})}{(0.0821 \text{ L·atm/(K·mol)}) \times (300 \text{ K})} \]
- Calculating gives:
\[ n \approx 0.1 \text{ moles} \]

7. **Calculate the Mass of Ozone:**
- The molar mass of ozone (O3) is:
\[ 3 \times 16 \text{ g/mol} = 48 \text{ g/mol} \]
- Therefore, the mass of ozone (m) is:
\[ m = n \times \text{molar mass} = 0.1 \text{ moles} \times 48 \text{ g/mol} = 4.8 \text{ g} \]

### Final Answer:
The mass of ozone dissolved in 400 g of water at 4.0 atm and 300 K is **4.8 g** (option d).