Question

The effusion rate of an unknown gas is measured and found to be 31.50 mL/min. Under identical experimental conditions, the effusion rate of O2 is found to be 30.50 mL/min. If the choices are CH4, CO, NO, CO2, and NO2, what is the identity of the unknown gas

Answer 1

Answer : The unknown gas is, NO

Solution :

According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

`R\propto \sqrt{(1)/(M)}`

or,

`((R_2)/(R_1))=\sqrt{(M_1)/(M_2)}` ..........(1)

where,

`R_1` = rate of effusion of unknown gas = 31.50 mL/min

`R_2` = rate of effusion of
`O_2` gas = 30.50 mL/min

`M_1` = molar mass of unknown gas = ?

`M_2` = molar mass of
`O_2` gas = 32 g/mole

Now put all the given values in the above formula 1, we get:

`((30.50mL/min)/(31.50mL/min))=\sqrt{(M_1)/(32g/mole)}`

`M_1=30.0g/mol`

From the this we conclude that the unknown gas is, NO that has 30.0 g/mol molar mass.

Hence, the unknown gas is, NO