Question #60299

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Answer 1 · 2015-11-15T00:31:58

$146 g / mol$ $146"g""/""mol"$

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

$R \propto \frac{1}{\sqrt{M_{r}}} = \frac{k}{\sqrt{M_{r}}}$ $Rprop(1)/sqrt(M_r)=k/sqrt(M_r)$

For 2 gases we can divide to get:

$\frac{R_{1}}{R_{2}} = \sqrt{\frac{M_{2}}{M_{1}}}$ $R_1/R_2=sqrt((M_2)/(M_1))$

We can express rate as $\frac{V}{t}$ $V/t$ so:

$\frac{V_{1}}{t} / \frac{V_{2}}{t} = \sqrt{\frac{M_{2}}{M_{1}}}$ $(V_1)/(t)/(V_2)/(t)=sqrt((M_2)/(M_1))$

Since $t$ $t$ 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 $rArr$

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

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

$:.0.274=40/M_1$

$:.M_1=40/0.274=146$