Asample of he gas (2.0 mmol) effused through a pinhole in 53 s. the same amount of an unknown gas, under the same conditions, effused through the pinhole in 248 s. the molecular mass of the unknown gas is g/mol.

Molecular mass of unknown gas is 87.58 grams per mol.

Explanation:

Let x be the molecular mass of the unknown gas.

Rate of effusion of a gas and its molecular mass is related as:

r ∝

where r is rate and M is molecular mass.

Molecular mass of He gas = 4 g/mol

Given: Same amount of gas is effused and at same conditions.

Let us say V mL of gas effused.

Then rate = V/t where t is time taken for effusion.

Since same amount of gases are effused V doesn't matter.

We can say:

t ∝ √M ⇒

then

Hence molecular mass of unknown gas is 87.58 grams per mol.

a gas, having a molecular weight of 13 kg/kmol and a kinematic viscosity of 0.25 cm2/s, flows through a pipe 0.25 m internal diameter and 5 km long at the rate of 0.4 m3/s and is delivered at atmospheric pressure.