The weight of 350 ml of diatomic gas at 0°C and 2 atm pressure is 1 g

1.	The weight of 350 ml of diatomic gas at 0°C and 2 atm pressure is 1 g. What is atomic mass of it
Answer» 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

The weight of 350 ml of diatomic gas at 0°C and 2 atm pressure is 1 g. What is atomic mass of it

Answer»

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

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