300 ~mL of a gas ' x ' of molar mass 32 ~g ~mol ⁻¹ is effused in 25 seconds. What volume of methane would effuse in the same time?

Step-by-step solution

Let, the volume of methane effused in 25 ~s be x ~mL . Now, from the Graham's law of diffusion, Rate of effusion 1 Molecular mass ...(i) Also, rate of effusion = Volume diffused Time taken ...(ii) From Eq. (i) and (ii) we get, aligned & r_x r_ CH ₄ = M_ CH ₄ M_x & 300 25 x 25 = 16 32 300 x = 1 2 aligned x=300 2 =424.26 ~mL 424 ~mL