Question

Question #60299

Answer 1

`146"g""/""mol"`

Explanation:

Graham's Law states that the rate of effusion is inversely proportional to the square root of the molar mass:

`Rprop(1)/sqrt(M_r)=k/sqrt(M_r)`

For 2 gases we can divide to get:

`R_1/R_2=sqrt((M_2)/(M_1))`

We can express rate as `V/t` so:

`(V_1)/(t)/(V_2)/(t)=sqrt((M_2)/(M_1))`

Since `t` is the same we can write:

`:.(V_1)/(cancel(t)).(cancel(t))/(V_2)=sqrt((M_2)/(M_1))`

`:.(V_1)/(V_2)=sqrt((M_2)/(M_1))`

I'll refer to the unknown gas as gas 1 and assume the `M_r` of Argon to be 40 `rArr`

`4.83/9.23=sqrt((40)/(M_1))`

`:.((4.83)/(9.23))^(2)=40/M_1`

`:.0.274=40/M_1`

`:.M_1=40/0.274=146`