Saved Bookmarks
| 1. |
N2 and A2 gas at 300 K are leaked into a vesselof 3 L for 10 minute from two identical holes. Theresulting pressure of the vessel becomes 6 atmand the mixture contains 0.4 mole of N2. Themolecular mass of A, is(1) 64.42(2) 22.4(3) 34.78(4) 41.14 |
|
Answer» 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
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. |
|