Question

An unknown monatomic gas diffuses twice as fast as Br2. What is the identity of this unknown gas?

Answer 1

Answer : The unknown monoatomic gas is, argon.

Explanation :

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 = 2R

`R_2` = rate of effusion of
`Br_2` gas = R

`M_1` = molar mass of unknown gas = ?

`M_2` = molar mass of
`Br_2` gas = 160 g/mole

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

`((R)/(2R))=\sqrt{(M_1)/(160g/mole)}`

`M_1=40g/mol`

From the molar mass 40 g/mol we conclude that the unknown monoatomic gas will be, argon.

Therefore, the unknown monoatomic gas is, argon.