Question

Under certain conditions Argon gas diffuses at a rate of 3.2 cm per second under the same conditions an unknown gas diffuses at the at a rate of 4.5 cm per second what is the approximate molar mass of the unknown gas

Answer 1

20 g/mol

Step-by-step explanation:

We can use Graham’s Law of diffusion:

The rate of diffusion (r) of a gas is inversely proportional to the square root of its molar mass (M).

`r = (1 )/(√(M))`

If you have two gases, the ratio of their rates of diffusion is

`(r_(2))/(r_(1)) = \sqrt{(M_(1))/(M_(2))}`

Squaring both sides, we get

`((r_(2))/(r_(1)))^(2) = (M_(1))/(M_(2))`

Solve for M₂:

`M_(2) = M_(1) * ((r_(1))/(r_(2)))^(2)`

`M_(2) = \text{39.95 g/mol} * (\frac{\text{3.2 cm/s}}{\text{4.5 cm/s}})^(2)= \text{39.95 g/mol} * (0.711 )^(2)`

`= \text{39.95 g/mol} * 0.506 = \textbf{20 g/mol}`