If the number of molecules of `SO_(2)` (atomic weight=64) effusing through an orifice of unit area of cross-section in unit time at `0^(@)C` and 1 atm pressure in n. the number of He molecules (atomic weight=4) effusin under similar conditions at `273^(@)C` and 0.25 atm is:

Options

A. `(n)/(sqrt(2))`
B. `nsqrt(2)`
C. `2n`
D. `(n)/(2)`

Answer

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. The formula can be expressed as follows:

\[ \frac{R_1}{R_2} = \frac{P_1}{P_2} \cdot \sqrt{\frac{m_2}{m_1} \cdot \frac{T_1}{T_2}} \]

Where:
- \( R_1 \) and \( R_2 \) are the rates of effusion of gases 1 and 2.
- \( P_1 \) and \( P_2 \) are the pressures of gases 1 and 2.
- \( m_1 \) and \( m_2 \) are the molar masses of gases 1 and 2.
- \( T_1 \) and \( T_2 \) are the temperatures of gases 1 and 2 in Kelvin.

### Step-by-Step Solution:

1. **Identify the gases and their properties**:
- Gas 1: \( SO_2 \)
- Molar mass (\( m_1 \)) = 64 g/mol
- Temperature (\( T_1 \)) = 0°C = 273 K
- Pressure (\( P_1 \)) = 1 atm
- Rate of effusion (\( R_1 \)) corresponds to \( n \).
- Gas 2: \( He \)
- Molar mass (\( m_2 \)) = 4 g/mol
- Temperature (\( T_2 \)) = 273°C = 546 K (273 + 273)
- Pressure (\( P_2 \)) = 0.25 atm
- Rate of effusion (\( R_2 \)) corresponds to \( n_2 \) (which we need to find).

2. **Set up the equation using Graham's law**:
\[ \frac{n}{n_2} = \frac{P_1}{P_2} \cdot \sqrt{\frac{m_2}{m_1} \cdot \frac{T_1}{T_2}} \]

3. **Substitute the known values into the equation**:
\[ \frac{n}{n_2} = \frac{1 \text{ atm}}{0.25 \text{ atm}} \cdot \sqrt{\frac{4 \text{ g/mol}}{64 \text{ g/mol}} \cdot \frac{273 \text{ K}}{546 \text{ K}}} \]

4. **Simplify the fractions**:
- \(\frac{1}{0.25} = 4\)
- \(\frac{4}{64} = \frac{1}{16}\)
- \(\frac{273}{546} = \frac{1}{2}\)

5. **Combine and simplify the square root**:
\[ \sqrt{\frac{1}{16} \cdot \frac{1}{2}} = \sqrt{\frac{1}{32}} = \frac{1}{\sqrt{32}} = \frac{1}{4\sqrt{2}} \]

6. **Final equation**:
\[ \frac{n}{n_2} = 4 \cdot \frac{1}{4\sqrt{2}} = \frac{1}{\sqrt{2}} \]

7. **Rearranging to find \( n_2 \)**:
\[ n_2 = n \cdot \sqrt{2} \]

### Conclusion:
The number of helium molecules effusing under the given conditions is:
\[ n_2 = \frac{n}{\sqrt{2}} \]