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

Answer:

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

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).

$r \propto \dfrac{1}{\sqrt{M}}$

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

$\dfrac{r_{2}}{r_{1}} = \sqrt{\dfrac{M_{1}}{M_{2}}}$

The time for diffusion is inversely proportional to the rate.

$\dfrac{t_{2}}{t_{1}} = \sqrt{\dfrac{M_{2}}{M_{1}}}$

Let CO₂ be Gas 1 and O₂ be Gas 2

Data:

M₁ = 44.01

M₂ = 32.00

Calculation

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