Question

N2 and A2 gas at 300 K are leaked into a vessel
of 3 L for 10 minute from two identical holes. The
resulting pressure of the vessel becomes 6 atm
and the mixture contains 0.4 mole of N2. The
molecular mass of A, is
(1) 64.42
(2) 22.4
(3) 34.78
(4) 41.14​

Answer 1

Answer:

The molecular mass of A is 41.14 g/ mol.

Explanation:

For the mixture of gas in the vessel, using the Ideal Gas Law

PV = nRT

where, P = pressure = 6 atm, V = volume = 3 L, T = temperature = 300 K, R = ideal gas constant = 0.082057 L atm mol^-1 K^-1 and n = total no. of moles in the mixture  

Or, n = PV/RT = (6 * 3) / (0.082057 * 300) = 18 / 24.617 = 0.7312 mol

No. of moles of N2 in the mixture, n1 = 0.4 mol

∴ No.of moles of A2 in the mixture, n2 = 0.7312 – 0.4 = 0.331 mol

The rate of effusion of N2, r1 =  0.4 / 10 min = 0.04 mol/min

The rate of effusion of A2, r2 = 0.331 / 10 min = 0.033 mol/min

Let the molecular mass of N2 be “M1” = 28 g/mol and molecular mass of A2 be “M2”.

According to Graham’s Law of Diffusion, we have  

$\frac{r1}{r2}$= $\sqrt{\frac{M2}{M1}}$

or, (0.04)² / (0.033)² = M2 / 28

or, M2 = (28 * 0.0016) / (0.001089) = 0.0448 / 0.001089 =  41.14 g/mol

Hence, the molecular mass of A i.e., M2 = 41.14 g/mol.