Answer: ethylene [tex](C_2H_4)[/tex] and carbon monoxide [tex](CO)[/tex]
Explanation:
This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows:
[tex]\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}[/tex]
Thus gases having similar molar masses will have same rate of diffusion.
Molar mass of ethylene [tex](C_2H_4)[/tex] = 28 g/mol
Molar mass of ethane [tex](C_2H_6)[/tex] = 30 g/mol
Molar mass of carbon monoxide [tex](CO)[/tex] = 28 g/mol
Molar mass of carbon dioxide [tex](CO_2)[/tex] = 44 g/mol
As ethylene and carbon monoxide have same molar masses , they have same rate of diffusion.
