Question

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

Answer 1

The two gases are
`Cl_(2)` and
`SF_(6)`

Step-by-step explanation:

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

`(r_(1))/(r_(2))=\sqrt{(M_(2))/(M_(1))}`

Where
`r_(1)` and
`r_(2)` are rate of effusion of gas-1 and gas-2 respectively.
`M_(1)` and
`M_(2)` are molar mass of gas-1 and gas-2 respectively.

Gas Molar mass (g/mol)

`O_(2)` 32

`Cl_(2)` 71

CO 28

`CO_(2)` 44

`SF_(6)` 146

So, clearly molar mass of
`SF_(6)` is greater than molar mass of
`Cl_(2)` . Hence
`Cl_(2)` will effuse faster.

Also,
`\sqrt{\frac{M_{SF_(6)}}{M_{Cl_(2)}}}=\sqrt{(146)/(71)}=1.43`

Hence, option (d) is correct.