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A mixture of two gases was allowed to effuse from a container. One of the gases escaped from the container 1.43 times as fast as the other one. The two gases could have been:a) CO and SF6
b) O2 and Cl2
c) CO and CO2 d) Cl2 and SF6
e) O2 and SF6

Respuesta :

Answer:

The two gases are [tex]Cl_{2}[/tex] and  [tex]SF_{6}[/tex]

Explanation:

According to Graham law of effusion for a binary mixture of gases-

                       [tex]\frac{r_{1}}{r_{2}}=\sqrt{\frac{M_{2}}{M_{1}}}[/tex]

Where [tex]r_{1}[/tex] and [tex]r_{2}[/tex] are rate of effusion of gas-1 and gas-2 respectively. [tex]M_{1}[/tex] and [tex]M_{2}[/tex] are molar mass of gas-1 and gas-2 respectively.

                             Gas                               Molar mass (g/mol)

                                     [tex]O_{2}[/tex]                            32

                                     [tex]Cl_{2}[/tex]                            71

                                   CO                                         28

                                     [tex]CO_{2}[/tex]                           44

                                     [tex]SF_{6}[/tex]                            146

So, clearly molar mass of [tex]SF_{6}[/tex] is greater than molar mass of [tex]Cl_{2}[/tex] . Hence [tex]Cl_{2}[/tex] will effuse faster.

Also, [tex]\sqrt{\frac{M_{SF_{6}}}{M_{Cl_{2}}}}=\sqrt{\frac{146}{71}}=1.43[/tex]

Hence, option (d) is correct.

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