Question

A sample of O2 gas (2.0 mmol) effused through a pinhole in 5.0 s. It will take __________ s for the same amount of CO2 to effuse under the same conditions.

Group of answer choices

Answer 1

`\large \boxed{\text{5.9 s}}`

Step-by-step explanation:

Graham’s Law applies to the effusion of gases:

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

`r \propto (1)/(√(M))`

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

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

The time for diffusion is inversely proportional to the rate.

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

Let CO₂ be Gas 1 and O₂ be Gas 2

Data:

M₁ = 44.01

M₂ = 32.00

Calculation

`\begin{array}{rcl}(t_(2))/(t_(1)) & = & \sqrt{(M_(2))/(M_(1))}\\\\\frac{t_(2)}{\text{5 s}}& = & \sqrt{(44.01)/(32.00)}\\\\& = & √(1.375)\\t_(2)& = & \text{5 s}* 1.173\\& = & \mathbf{5.9 s} \\\end{array}\\\text{It will take $\large \boxed{\textbf{5.9 s}}$ for the carbon dioxide to effuse.}`