Question

Carbon dioxide gas (CO2) effuses 3.2 times faster than an unknown gas. Determine the molar mass of the unknown gas.

Answer 1

To solve this exercise, we are going to use Graham's Law of diffusion:

`\frac{rate1}{\text{rate}2}=\sqrt[]{(M2)/(M1)}`

1 is for the rate and molar mass (M) of the first gas.

2 is for the second gas.

Data:

CO2 will be gas number 1

M1 = 44 g/mol

rate 1

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Unknown gas is number 2

rate 2 and M2

rate 1 = 3.2 x rate 2

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From Graham's law we clear M2:

`\begin{gathered} (\frac{rate1}{\text{rate}2})^2xM1\text{ = M2} \\ (\frac{3.2rate2}{\text{rate}2})^2x44\frac{g}{\text{mol}}=\text{ M2} \\ 450\text{ g/mol = M2} \end{gathered}`

Answer: 450 g/mol