80 mL of `O_(2)` takes 2 minutes to pass through the hole. What volume of `SO_(2)` will pass through the hole in 3 minute?

Options

A. `(120)/(sqrt(2))` mL
B. `120xxsqrt(2)` mL
C. `(12)/(sqrt(2))` mL
D. None of these

Explanation

To solve the problem, we will use Graham's law of effusion, which states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. We can set up the relationship between the rates of diffusion of oxygen (O₂) and sulfur dioxide (SO₂) using the formula:

\[ \frac{R_1}{R_2} = \frac{V_1/T_1}{V_2/T_2} = \sqrt{\frac{M_2}{M_1}} \]

Where:
- \(R_1\) and \(R_2\) are the rates of diffusion for O₂ and SO₂, respectively.
- \(V_1\) and \(V_2\) are the volumes of O₂ and SO₂, respectively.
- \(T_1\) and \(T_2\) are the times taken for O₂ and SO₂, respectively.
- \(M_1\) and \(M_2\) are the molar masses of O₂ and SO₂, respectively.

### Step 1: Identify the known values
- Volume of O₂, \(V_1 = 80 \, \text{mL}\)
- Time for O₂, \(T_1 = 2 \, \text{minutes}\)
- Time for SO₂, \(T_2 = 3 \, \text{minutes}\)
- Molar mass of O₂, \(M_1 = 32 \, \text{g/mol}\)
- Molar mass of SO₂, \(M_2 = 64 \, \text{g/mol}\)

### Step 2: Set up the equation using Graham's law
Using the relationship derived from Graham's law, we can write:

\[ \frac{80 \, \text{mL}/2 \, \text{min}}{V_2/3 \, \text{min}} = \sqrt{\frac{64}{32}} \]

### Step 3: Simplify the equation
The right side simplifies as follows:

\[ \sqrt{\frac{64}{32}} = \sqrt{2} \]

Now, substituting this back into our equation gives:

\[ \frac{80/2}{V_2/3} = \sqrt{2} \]

### Step 4: Cross-multiply and solve for \(V_2\)
Cross-multiplying gives:

\[ 80 \cdot 3 = 2 \cdot V_2 \cdot \sqrt{2} \]

This simplifies to:

\[ 240 = 2V_2\sqrt{2} \]

Now, divide both sides by \(2\sqrt{2}\):

\[ V_2 = \frac{240}{2\sqrt{2}} = \frac{120}{\sqrt{2}} \, \text{mL} \]

### Step 5: Rationalize the denominator (optional)
To express \(V_2\) in a more standard form, we can rationalize the denominator:

\[ V_2 = \frac{120\sqrt{2}}{2} = 60\sqrt{2} \, \text{mL} \]

### Final Answer
The volume of SO₂ that will pass through the hole in 3 minutes is:

\[ V_2 = \frac{120}{\sqrt{2}} \, \text{mL} \quad \text{or} \quad 60\sqrt{2} \, \text{mL} \]