360 cm qube of methane gas diffused through a porous membrane in 15 mi

1.	360 cm qube of methane gas diffused through a porous membrane in 15 minutes. under similar conditions, 120 cm qube of another gas diffused in 10 minutes. find the molar mass of the gas.
Answer» Answer: Solution Rate of diffusion of methane r1=Volume of METHAN eTime of dif fusion of methan e =360cm315min=24cm3min−1 Molar mass of CH4(M1)=16gmol−1 Unknown GAS Rate of diffusion of unknown gas r2=Volume of unknown gasTime of dif fusion =120cm310min=12cm3min−1 Molar mass of unknown gas (M2)=? Acording of Graham's law of diffusion, r1r2=M2M1−−−−√ 24cm3min−112cm3min−1=M216gmol−2−−−−−−−−√ Squaring of both SIDES, M2=24×24×1612×12=64 Molar mass of the unknown gas =64g.mol−1 Explanation:

360 cm qube of methane gas diffused through a porous membrane in 15 minutes. under similar conditions, 120 cm qube of another gas diffused in 10 minutes. find the molar mass of the gas.

Answer»

Answer:

Solution

Rate of diffusion of methane

r1=Volume of METHAN eTime of dif fusion of methan e

=360cm315min=24cm3min−1

Molar mass of CH4(M1)=16gmol−1

Unknown GAS

Rate of diffusion of unknown gas

r2=Volume of unknown gasTime of dif fusion

=120cm310min=12cm3min−1

Molar mass of unknown gas (M2)=?

Acording of Graham's law of diffusion,

r1r2=M2M1−−−−√

24cm3min−112cm3min−1=M216gmol−2−−−−−−−−√

Squaring of both SIDES,

M2=24×24×1612×12=64

Molar mass of the unknown gas =64g.mol−1

Explanation: