A quantity of gas is collected in a graduated tube over the mercury. The volume of gas at `18^(@)C` is 50 mL and the level of mercuty in the tube is 100mm above the outside mercuty level. The barometer reads 750 torr. Hence, volume of gas 1 atm and `0^(@)C` is approximately:

Options

A. `22 ml`
B. `40 ml`
C. `20ml`
D. `44ml`

Explanation

To solve the problem, we will use the Ideal Gas Law, which states that:

\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Volume of gas at \( T_1 = 18^\circ C \) is \( V_1 = 50 \, \text{mL} \).
- The barometer reads \( 750 \, \text{torr} \).
- The mercury level in the tube is \( 100 \, \text{mm} \) above the outside mercury level.
- Convert \( T_1 \) to Kelvin:
\[ T_1 = 18 + 273 = 291 \, \text{K} \]

2. **Calculate the Pressure of the Gas \( P_1 \):**
- The pressure of the gas \( P_1 \) can be calculated as:
\[ P_1 = \text{Barometric Pressure} - \text{Height of Mercury Column} \]
\[ P_1 = 750 \, \text{torr} - 100 \, \text{mm} = 650 \, \text{torr} \]

3. **Identify the Conditions for \( P_2 \) and \( T_2 \):**
- We need to find the volume \( V_2 \) at:
- \( P_2 = 760 \, \text{torr} \) (1 atm)
- \( T_2 = 0^\circ C = 273 \, \text{K} \)

4. **Set Up the Ideal Gas Law Equation:**
- Substitute the known values into the equation:
\[ \frac{650 \, \text{torr} \times 50 \, \text{mL}}{291 \, \text{K}} = \frac{760 \, \text{torr} \times V_2}{273 \, \text{K}} \]

5. **Cross-Multiply to Solve for \( V_2 \):**
- Rearranging gives:
\[ V_2 = \frac{650 \times 50 \times 273}{760 \times 291} \]

6. **Calculate \( V_2 \):**
- Performing the calculations:
\[ V_2 = \frac{650 \times 50 \times 273}{760 \times 291} \approx 40.0 \, \text{mL} \]

### Final Answer:
The volume of gas at \( 1 \, \text{atm} \) and \( 0^\circ C \) is approximately \( 40 \, \text{mL} \).