In which one of the following, does the given amount of chlorine exert the least pressure in a vessel of capacity 1 d `m^(3)` at 273 K ?

Options

A. `0.0355` g
B. `0.071` g
C. `6.023xx10^(21)` molecules
D. `0.02` mol

Explanation

To determine in which case the given amount of chlorine exerts the least pressure in a vessel of capacity 1 dm³ at 273 K, we will use the Ideal Gas Law, which is given by the equation:

\[ PV = nRT \]

Where:
- \( P \) = pressure
- \( V \) = volume
- \( n \) = number of moles
- \( R \) = ideal gas constant (0.0821 L·atm/(K·mol) or 8.314 J/(K·mol))
- \( T \) = temperature in Kelvin

Since the volume (1 dm³ = 1 L) and temperature (273 K) are constant, the pressure exerted by the gas is directly proportional to the number of moles of chlorine gas present. Therefore, we need to calculate the number of moles for each option provided and find the one with the least number of moles.

### Step-by-step Solution:

1. **Calculate Moles for Each Option:**
- **Option 1:** Given mass = 0.0355 g
\[ n_1 = \frac{0.0355 \text{ g}}{71 \text{ g/mol}} = 5 \times 10^{-4} \text{ moles} \]

- **Option 2:** Given mass = 0.071 g
\[ n_2 = \frac{0.071 \text{ g}}{71 \text{ g/mol}} = 10 \times 10^{-3} \text{ moles} = 10^{-2} \text{ moles} \]

- **Option 3:** Given number of molecules = \( 6.023 \times 10^{21} \)
\[ n_3 = \frac{6.023 \times 10^{21}}{6.023 \times 10^{23}} = 10^{-2} \text{ moles} \]

- **Option 4:** Given moles = 0.02 moles
\[ n_4 = 0.02 \text{ moles} \]

2. **Compare the Number of Moles:**
- \( n_1 = 5 \times 10^{-4} \)
- \( n_2 = 10^{-2} \)
- \( n_3 = 10^{-2} \)
- \( n_4 = 0.02 \)

3. **Identify the Least Number of Moles:**
- The least number of moles is \( n_1 = 5 \times 10^{-4} \).

4. **Conclusion:**
- Since pressure is directly proportional to the number of moles, the option with the least number of moles will exert the least pressure. Therefore, the correct answer is **Option 1**.