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.

the atomic number will stay the same no matter how many electrons it has, so it doesn’t decrease nor increase. it’ll only change with a different number of protons