**Solution:**

Use the formula PV = nRT

P = 2 atm

V = 350 ml. = 0.35 L

R = 0.0821 L atm K^{-1} mol^{-1}

T = 0º C = 273 K

Putting the values in given formula we get :

n = PV/RT

= 2 atm x 0.35 L/ 0.0821 L atm K^{-1} mol^{-1} x 273 K

= 0.0312 mol

Now weight of 0.0312 mol of gas is 1 gram

therefore weight of 1 mol of gas would be 1/0.0312 = 32 g mol^{-1}

Since the gas is diatomic therefore weight of 1 mole atom = 16 g mol^{-1}

Weight of 1 atom of gas = 16 g mol^{-1}/ 6.022 x 1023 atom mol^{-1}

= 2.65 x 10^{-23} g /atom