Question

An unknown gas is found to diffuse through a porous membrane 4.11 times more slowly than h2 what is the molecular weight of the gas?

Answer 1

To answer this question, you need to know Graham's Law of Effusion/Diffusion formula. In this formula, the rate of diffusion/effusion would be influenced by the mass. As the molecule has bigger mass, the rate should be slower because it will be harder to pass the membrane. The calculation should be:
Rate 1 / Rate 2 = √[M2/M1] 
4.11/1= √[M2/2] 
M2=33.78 g/mol

Answer 2

Answer:

33.78 g/mol

Explanation:

Diffusion is the transportation of the molecules by their movements from the fluid. It can be characterized by Graham's Law, which states that the rate of diffusion is inversely proportional to the square of the molar mass. The rate is also the amount divided bt the time of diffusion.

Calling the amount x, the time t and the molar mass M, and H2 as the substance 1, the ratio between the rates of the gases is:

(x1/t1)/(x2/t2) = (√M2)/(√M1)

For the same amount of gas, x1 = x2, the gas 2 is 4.11 times slowly, so t2 = 4.11t1 (if it slowly it will take longer to diffuse). The molar mass of H2 is 2 g/mol. Thus:

t2/t1 = (√M2)/√2

4.11t1/t1 = (√M2)/√2

4.11 = (√M2)/√2

√M2 = 4.11*√2

√M2 = 5.8124

M2 = (5.8124)²

M2 = 33.78 g/mol