Question

A sample of gas is observed to effuse through a pourous barrier in 4.98 minutes. Under the same conditions, the same number of moles of an unknown gas requires 6.34 minutes to effuse through the same barrier. The molar mass of the unknown gas is:________.
g/mol.

Answer 1

Answer:

The molar mass of the unknown gas is $\mathbf{ 51.865 \ g/mol}$

Explanation:

Let assume that  the gas is  O2 gas

O2 gas is to effuse through a porous barrier in time t₁ = 4.98 minutes.

Under the same conditions;

the same number of moles of an unknown gas requires  time t₂  =  6.34 minutes to effuse through the same barrier.

From Graham's Law of Diffusion;

Graham's Law of Diffusion states that, at a constant temperature and pressure; the rate of diffusion of a gas is inversely proportional to the square root of its density.

i.e

$R \ \alpha \ \dfrac{1}{\sqrt{d}}$

$R = \dfrac{k}{d}$  where K = constant

If we compare the rate o diffusion of two gases;

$\dfrac{R_1}{R_2}= {\sqrt{\dfrac{d_2}{d_1}}$

Since the density of a gas d is proportional to its relative molecular mass M. Then;

$\dfrac{R_1}{R_2}= {\sqrt{\dfrac{M_2}{M_1}}$

Rate is the reciprocal of time ; i.e

$R = \dfrac{1}{t}$

Thus; replacing the value of R into the above previous equation;we have:

$\dfrac{R_1}{R_2}={\dfrac{t_2}{t_1}}$

We can equally say:

${\dfrac{t_2}{t_1}}= {\sqrt{\dfrac{M_2}{M_1}}$

${\dfrac{6.34}{4.98}}= {\sqrt{\dfrac{M_2}{32}}$

$M_2 = 32 \times ( \dfrac{6.34}{4.98})^2$

$M_2 = 32 \times ( 1.273092369)^2$

$M_2 = 32 \times 1.62076418$

$\mathbf{M_2 = 51.865 \ g/mol}$