Answer to Question #145409 in General Chemistry for Rien Kin

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

Let x be the molecular mass of the unknown gas.

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

r ∝ $\frac{1}{M}$

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 = 53sec

t_{unknown gas} = 248sec

r = $\frac{V}{t}$ = $\frac{k}{\sqrt M}$

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

We can say:

t ∝ √M ⇒

then

$\frac{53}{248}$ = $\frac{\sqrt4}{\sqrt x}$

x = 87.58 g/mol

or we can say 88g/mol

Hence molecular mass of unknown gas is 88 grams per mol.