Question

For 10 minute each, at 0 °C, from two identical holes nitrogen and an unknown gas are leaked into a common
vessel of 4 litre capacity. The resulting pressure is 2.8 atm and the mixture contains
0.4 mole of nitrogen. What is the molar mass of unknown gas?​

Answer 1

Answer:

Temperature, T = 0 °C = 273 K

Volume, V = 4 litre

Pressure, P = 2.8 atm

Ideal gas constant, R = 0.0821 L atm mol⁻¹ K⁻¹

Let the no. of mole of N2 be “n1” = 0.4 mole and the unknown gas be “n2”.

∴ Total moles, n = n1 + n2 = 0.4 + n2 …. (i)

Using Ideal Gas Law, we get

PV = nRT

∴ n = PV / RT  

⇒ 0.4 + n2 = (2.8 * 4) / (0.0821 * 273)

⇒ 0.4 + n2 = 11.2 / 22.41

⇒ n2 = 0.499 – 0.4 = 0.099 ≈ 0.1 mole

Let the rate of diffusion of N2 be “r1” and unknown gas be “r2”.  

Let the molar mass of N2 be “M1” = 28 g/mol and the molar mass of unknown gas be “M2”.

Time taken for diffusion of each gas = 10 minute

Using Graham’s Law,

r1/r2 = √[M2/M1]

⇒ [n1/10] / [n2/10] = √[M2/M1]

⇒ [0.4/10] * [10/0.1] = √[M2/28]

⇒ [0.4/0.1]^2 = M2 / 28

⇒ 16 = M2/28

⇒ M2 = 16 * 28 = 448 g/mol

Hence, the molar mass of unknown gas is 448 g/mol.