Question

Is 150 ml of carbon monoxide effused in 25 seconds what volume of methane would if used in same time?

Answer 1

Answer:

212.132 mL

Explanation:

According to the Graham's law of the diffusion, the rate of the effusion of the gas is inversely proportional to the square root of the molar mass of that gas.

The rate of effusion is defined as the amount of gas transferred per unit time.

$\frac{rate of effusion of carbon monoxide}{rate of effusion of methane} =\frac{\sqrt{molar mass of methane} }{\sqrt{molar mass of carbon monoxide} }$

The above equation can be expressed in the volume of gas transferred per unit time.

\frac{\frac{volume of the carbon monoxide effused}{time}}{\frac{volume of the methane effused}{time}} =\frac{\sqrt{molar mass of methane} }{\sqrt{molar mass of carbon monoxide} }

The molar mass of  carbon monoxide is  $28.0 g.mol^{-1}$.

The molar mass of  methane is  $16.0 g.mol^{-1}$.

volume of the carbon monoxide effused is  150 mL.

Substitute the value in the above expression.

\frac{\frac{150 mL}{25 sec}}{\frac{volume of the methane effused}{25 sec}} =\frac{\sqrt{14.0 g.mol^{-1}} }{\sqrt{28.0 g.mol^{-1}} }

$volume of the methane effused=\frac{\sqrt{28 gmol^{-1}} }{\sqrt{14 gmol^{-1}} }(150 mL)$

= 212.132 mL