Question

8. Neon gas is found to effuse through a hole. Careful measurement indicates that 1.42 x 10 mol effused in 5.8 minutes. How long (in min) would it take for the same amount of xenon gas to effuse through the same hole under the same conditions? A: 14.8 RKSHEET 9 9. A 2.00 L bottle contains the following gases: 0.0150 mol He, 0.100 mol Ny, and 0.00250 mol Xe. If the temperature of the gas mixture is 22.4 °C, what is the partial pressure (in atm) of each gas? What is the total pressure (in atm) in the bottle? A: 0.182, 1.21; 3.03 x 10-?; 1.42

Answer 1

According to the Graham's law of diffusion, the rate of effusion of the gases is inversely proportional to the square root of their molar mass.

$Rate=\sqrt{\frac{1}{M}}$

Here M is the molar mass of the compound.

If R₁ is the rate of diffusion of neon gas and R₂ is that of xenon.

And M₁ is the molar mass of neon(20.17 g/mol) and M₂ is that of xenon(131.29 g/mol), then,

$\frac{R_{1}}{R_{2}}=\sqrt{\frac{M_{2}}{M_{1}}}$

$\frac{5.8 min}{R_{2}}=\sqrt{\frac{131.29 g/mol}{20.17g/mol}}$

$R_{2}=2.27 min$

Thus the rate of effusion of xenon gas is calculated to be 2.27 min.