Question

A sample 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.

Answer 1

Molecular mass of unknown gas is 87.58 grams per mol.

Step-by-step 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 ∝
`(1)/(√(M) )`

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.

`t_(He)=53\ s\\t_(unknown\ gas)=248\ s`

`r=(V)/(t)=(k)/(√(M) )`

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

We can say:

t ∝ √M ⇒

then

`(53)/(248)=(√(4) )/(√(x) )\\x=87.58\ g\ per\ mol`

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