Question

1.538×10−4mol of an unidentified gaseous substance effuses through a

tiny hole in 89.7 s. Under identical conditions, 1.653×10−4mol of argon

gas takes 87.9 s to effuse.

a. What is the molar mass of the unidentified substance (in g/mol)?​

Answer 1

Given info : 1.538 × 10¯⁴ mol of an unidentified gaseous substance effuses through a tiny hole in 89.7 s. under identical conditions, 1.653 × 10¯⁴ mol of Argon take 87.9 sec to effuse.

To find : the molar mass of the unidentified substance in g/mol is...

solution : from Graham's law of diffusion/effusion, rate of diffusion/effusion is inversely proportional to square root of their density.

i.e., r ∝ 1/√d ..(1)

but we know density of an ideal gas is given by, d = PM/RT

so, d ∝ M ...(2)

from equations (1) and (2) we get,

then, r ∝ 1/√M

r = volume of effused gas/time taken

under identical conditions of temperature and pressure, volume of gas is directly proportional to no of moles

so, r = no of mole of effused gas/Time taken

here, r₁ = 1.538 × 10¯⁴ mol/89.7 sec

r₂ = 1.653 × 10¯⁴ mol/87.9 sec

M₂ = 40 amu

now, r₁/r₂ = √(M₂/M₁)

⇒(1.538 × 10¯⁴ mol/89.7 sec)/(1.653 × 10¯⁴ mol/87.9 sec) = √(40/M₁)

⇒1.538 × 87.9/(1.653 × 89.7) = √(40/M₁)

⇒M₁ = 48 amu

Therefore the molar mass of unidentified gas is 48 g/mol