The time taken for effusion of 64 mL of oxygen will be as the time taken for the effusion of which of the following gases under identical conditions ?

Options

A. `64 mL " of " H_(2)`
B. `100 mL " of " N_(2)`
C. `64 mL " of " CO_(2)`
D. `45.24 mL" of " SO_(2)`

Explanation

To solve the problem of determining which gas will take the same time to effuse as 64 mL of oxygen under identical conditions, we can use Graham's law of effusion. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand Graham's Law of Effusion
Graham's law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass (molecular weight). Mathematically, this can be expressed as:
\[ \text{Rate} \propto \frac{1}{\sqrt{M}} \]
where \( M \) is the molar mass of the gas.

### Step 2: Set Up the Equation
According to Graham's law, if we have two gases (Gas 1 and Gas 2), we can write:
\[ \frac{V_1}{t_1} = \frac{V_2}{t_2} \propto \frac{1}{\sqrt{M_1}} \quad \text{and} \quad \frac{1}{\sqrt{M_2}} \]
Where:
- \( V_1 \) and \( V_2 \) are the volumes of the gases.
- \( t_1 \) and \( t_2 \) are the times taken for effusion.
- \( M_1 \) and \( M_2 \) are the molar masses of the gases.

### Step 3: Apply the Given Values
Let’s denote:
- \( V_1 = 64 \) mL (volume of oxygen)
- \( M_1 = 32 \) g/mol (molar mass of oxygen, \( O_2 \))
- \( V_2 \) is the volume of the unknown gas
- \( M_2 \) is the molar mass of the unknown gas

From Graham's law, we can write:
\[ \frac{64}{t_{O_2}} = \frac{V_2}{t_{unknown}} \propto \frac{1}{\sqrt{32}} \quad \text{and} \quad \frac{1}{\sqrt{M_2}} \]

### Step 4: Solve for the Unknown Gas
Since we want the time taken for effusion to be the same for both gases, we can set the ratios equal:
\[ \frac{64}{t_{O_2}} = \frac{V_2}{t_{unknown}} \Rightarrow \frac{64}{V_2} = \frac{\sqrt{M_2}}{\sqrt{32}} \]
Squaring both sides gives:
\[ \frac{64^2}{V_2^2} = \frac{M_2}{32} \]
Rearranging gives:
\[ M_2 = \frac{64^2}{V_2^2} \times 32 \]

### Step 5: Calculate for Each Option
Now, we will evaluate the given options for \( V_2 \) and find \( M_2 \) for each:

1. **Option A: 64 mL of H2**
- \( M_2 = 2 \) g/mol (not equal to 32)

2. **Option B: 100 mL of unknown gas**
- \( M_2 = \frac{64^2}{100^2} \times 32 = \frac{4096}{10000} \times 32 = 13.12 \) g/mol (not equal to 32)

3. **Option C: 64 mL of CO2**
- \( M_2 = 44 \) g/mol (not equal to 32)

4. **Option D: 45.24 mL of SO2**
- \( M_2 = \frac{64^2}{45.24^2} \times 32 \)
- Calculating gives \( M_2 \approx 64 \) g/mol (which is very close to the molar mass of SO2)

### Conclusion
The gas that will take the same time to effuse as 64 mL of oxygen under identical conditions is **SO2** (Sulfur Dioxide), which has a molar mass of approximately 64 g/mol.