The weight of 350 ml a diatomic gas at `0^(@)C` and 2atm pressure is 1 gm.The weight of one atom is :-

Options

A. `(16)/(N_(A))`
B. `(32)/(N_(A))`
C. `16N_(A)`
D. `32N_(A)`

Explanation

To solve the problem, we need to find the weight of one atom of a diatomic gas given the weight of 350 ml of the gas at 0°C and 2 atm pressure is 1 gram.

### Step-by-Step Solution:

1. **Convert Given Volume to Liters**:
- The volume of the gas is given as 350 ml.
- To convert this to liters, we divide by 1000.
\[ V = \frac{350 \, \text{ml}}{1000} = 0.350 \, \text{L} \]

2. **Use the Ideal Gas Law**:
- The ideal gas equation is given by \( PV = nRT \).
- Here, \( P = 2 \, \text{atm} \), \( V = 0.350 \, \text{L} \), \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \), and \( T = 273 \, \text{K} \).
- Rearranging the equation to find \( n \) (number of moles):
\[ n = \frac{PV}{RT} \]
\[ n = \frac{(2 \, \text{atm})(0.350 \, \text{L})}{(0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1})(273 \, \text{K})} \]

3. **Calculate the Number of Moles**:
- Plugging in the values:
\[ n = \frac{0.700}{22.414} \approx 0.0312 \, \text{mol} \]

4. **Calculate Molar Mass**:
- Given that the weight of the gas is 1 gram, we can find the molar mass \( M \):
\[ M = \frac{\text{mass}}{n} = \frac{1 \, \text{g}}{0.0312 \, \text{mol}} \approx 32.05 \, \text{g/mol} \]

5. **Determine the Weight of One Molecule**:
- Since the gas is diatomic (let's denote it as \( X_2 \)), the molar mass of one molecule (which consists of 2 atoms) is:
\[ \text{Molar mass of } X_2 = 32 \, \text{g/mol} \]
- To find the weight of one atom, we first find the number of molecules in one mole:
\[ \text{Number of molecules in 1 mole} = 6.022 \times 10^{23} \]
- Therefore, the weight of one atom is:
\[ \text{Weight of one atom} = \frac{32 \, \text{g/mol}}{2 \times 6.022 \times 10^{23}} = \frac{32 \times 10^{-3} \, \text{g}}{2 \times 6.022 \times 10^{23}} \]
\[ = \frac{32 \times 10^{-3}}{12.044 \times 10^{23}} \approx 1.33 \times 10^{-25} \, \text{g} \]

### Final Answer:
The weight of one atom of the diatomic gas is approximately \( 1.33 \times 10^{-25} \, \text{g} \).