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.

$R\propto \sqrt{\frac{1}{M}}$

or,

$(\frac{R_2}{R_1})=\sqrt{\frac{M_1}{M_2}}$       ..........(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:

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

$M_1=40g/mol$

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

Therefore, the unknown monoatomic gas is, argon.