`O_(2)` and `SO_(2)` gases are filled in ratio of 1 : 3 by mass in a closed container of 3 L at temperature of `27^(@)C`. The partial pressure of `O_(2)` is 0.60 atm, the concentration of `SO_(2)` would be

Options

A. 0.36
B. 0.036
C. 3.6
D. 36

Explanation

To solve the problem, we need to find the concentration of \( SO_2 \) gas in a closed container. Let's break down the solution step by step.

### Step 1: Understand the given data
- The gases \( O_2 \) and \( SO_2 \) are filled in a ratio of 1:3 by mass.
- The volume of the container is 3 L.
- The temperature is \( 27^\circ C \) (which is \( 300 \, K \)).
- The partial pressure of \( O_2 \) is \( 0.60 \, atm \).

### Step 2: Calculate the total pressure
From the ideal gas law, we know that the total pressure \( P_t \) can be expressed in terms of the partial pressures:
\[ P_t = P_{O_2} + P_{SO_2} \]
Given \( P_{O_2} = 0.60 \, atm \), we need to find \( P_{SO_2} \).

### Step 3: Calculate the mole fraction of \( O_2 \)
The mole fraction of \( O_2 \) can be expressed as:
\[ P_{O_2} = \chi_{O_2} \cdot P_t \]
Where \( \chi_{O_2} \) is the mole fraction of \( O_2 \). We can rearrange this to find \( P_t \):
\[ 0.60 = \chi_{O_2} \cdot P_t \]

### Step 4: Determine the mass ratio and convert to moles
Let the mass of \( O_2 \) be \( x \) grams. Since the ratio of \( O_2 \) to \( SO_2 \) is 1:3 by mass, the mass of \( SO_2 \) will be \( 3x \) grams.

Now, we can convert these masses to moles:
- Moles of \( O_2 \):
\[ n_{O_2} = \frac{x}{32} \]
- Moles of \( SO_2 \):
\[ n_{SO_2} = \frac{3x}{64} \]

### Step 5: Calculate the mole fraction of \( O_2 \)
The total number of moles \( n_{total} \) is:
\[ n_{total} = n_{O_2} + n_{SO_2} = \frac{x}{32} + \frac{3x}{64} \]
To add these fractions, we need a common denominator:
\[ n_{total} = \frac{2x}{64} + \frac{3x}{64} = \frac{5x}{64} \]

Now, the mole fraction of \( O_2 \) is:
\[ \chi_{O_2} = \frac{n_{O_2}}{n_{total}} = \frac{\frac{x}{32}}{\frac{5x}{64}} = \frac{64}{160} = \frac{2}{5} \]

### Step 6: Calculate total pressure \( P_t \)
Substituting \( \chi_{O_2} \) back into the equation for \( P_t \):
\[ 0.60 = \frac{2}{5} \cdot P_t \]
Solving for \( P_t \):
\[ P_t = 0.60 \cdot \frac{5}{2} = 1.5 \, atm \]

### Step 7: Calculate the partial pressure of \( SO_2 \)
Now we can find the partial pressure of \( SO_2 \):
\[ P_{SO_2} = P_t - P_{O_2} = 1.5 - 0.60 = 0.90 \, atm \]

### Step 8: Calculate the concentration of \( SO_2 \)
Using the ideal gas law \( PV = nRT \), we can express concentration \( C \) as:
\[ C = \frac{P}{RT} \]
Where:
- \( P = 0.90 \, atm \)
- \( R = 0.0821 \, L \cdot atm/(K \cdot mol) \)
- \( T = 300 \, K \)

Substituting the values:
\[ C = \frac{0.90}{0.0821 \cdot 300} = \frac{0.90}{24.63} \approx 0.0365 \, mol/L \]

### Final Answer
The concentration of \( SO_2 \) is approximately \( 0.0365 \, mol/L \).

---