To solve this exercise, we are going to use Graham's Law of diffusion:
[tex]\frac{rate1}{\text{rate}2}=\sqrt[]{\frac{M2}{M1}}[/tex]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:
[tex]\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}[/tex]Answer: 450 g/mol
