Because of their different molar masses, ${\mathbf{\text{H}}}_{\mathbf{\text{2}}}$androle="math" localid="1663325761535" ${\mathit{D}}_{\mathbf{\text{2}}}$effuse at different rates (Section).

(a) If it takes$\mathbf{\text{16.5}}$min for$\mathbf{\text{0.10mol}}$of${\mathbf{\text{H}}}_{\mathbf{\text{2}}}$to effuse, how long does it take for$\mathbf{0}\mathbf{.}\mathbf{10}\mathbf{}\mathit{m}\mathit{o}\mathit{l}$of${\mathit{D}}_{\mathbf{\text{2}}}$to do so in the same apparatus at the sameand? (b) How many effusion steps does it take to separate an equimolar mixture of${\mathit{D}}_{\mathbf{\text{2}}}$and${\mathbf{\text{H}}}_{\mathbf{\text{2}}}$to$\mathbf{99}\mathit{m}\mathit{o}\mathit{l}\mathbf{\%}$purity?

It was discovered that a gas's rate of effusion is inversely proportional to the square root of its particles' molar mass.

Effusion is a process in which gas escapes through a small hole into the evacuated space of a container.

The square root of the given gas density determines the effusion rate of gas, which is defined as the number of molecules exiting per unit of time.

Continuing, because density is related to molar mass,

$\text{Graham'slawofeffusion:}\frac{{\text{rate}}_{\text{A}}}{{\text{rate}}_{\text{B}}}\text{=}\sqrt{\frac{{\text{M}}_{\text{R}}\text{(A)}}{{\text{M}}_{\text{R}}\text{(B)}}}$

The basic principle is that at given and the gas with a smaller molar mass effuses (escapes) faster because its molecules move faster and have a better chance of reaching the hole and escaping.

a)

It is known that: - for$H_2$, it takes16.5min to effuse$0.10\text{mol}$of gas; - for$D_2$,time to effuse is unknown for$0.10\text{mol}$of gas

Time to effuse in unknown for$D_{\mathbf{2}}$0.10mol of gas

The following dependency can be stated using Graham's law of effusion.

$\frac{\text{rate}\left({\text{H}}_{\text{2}}\right)}{\text{rate}\left({\text{D}}_{\text{2}}\right)}\text{=}\sqrt{\frac{{\text{M}}_{\text{R}}\left({\text{D}}_{\text{2}}\right)}{{\text{M}}_{\text{R}}\left({\text{H}}_{\text{2}}\right)}}$

Using the known values as substitutes,

$$\begin{gathered} \frac{rate\left(H_2\right)}{rate\left(D_2\right)}=\sqrt{\frac{4.028\text{g}/\text{mol}}{2.016\text{g}/\text{mol}}} \\ \frac{rate\left(H_2\right)}{rate\left(D_2\right)}=\sqrt{1.998} \\ \frac{rate\left({\mathbf{H}}_2\right)}{rate\left({\mathbf{D}}_2\right)}=\mathbf{1}.\mathbf{4135} \end{gathered}$$

As a result,$D_2$takes$\text{1.4135}$ more time to effuse than$H_2$ As a result, the time for to$D_2$ effuse is now.

$$\begin{gathered} Time\left(D_2\right)=1.4135\times 16.5\text{min} \\ Time\left(D_2\right)=23.323\text{min} \end{gathered}$$

The effusion takes $\text{23.323}$minutes to complete.

Therefore, the time required for effusion is $\text{23.323}$minutes.

b)

Assume the number of effusion stages is$x$.Given that the ultimate purity is$99.00\%$and the initial molar ratio is equimolar, i.e. the same quantity is present -$50\%{\text{H}}_2$and$50\%D_2$

The change in the purity is then

$\frac{\text{final}}{\text{initial}}=\frac{0.99}{0.50}=1.98$

Then there's the shift in concentration caused by the effusion:

$1.98=1.{4135}^x$

Using to re-write the equation, expressing $x$:

$$\begin{gathered} \log (1.98)=\log \left({1.4135}^x\right) \\ \log (1.98)=x·\log (1.4135) \end{gathered}$$

$$\begin{gathered} x=\frac{\log (1.98)}{\log (1.4135)} \\ x=1.974 \end{gathered}$$

About two effusion stages are necessary to obtain role="math" localid="1663326670398" $99\backslash \%$pure$H_2$ from an equimolar mixture.

Therefore, in order to receive 99% pure from equimolar mixture, about 2 effusion steps are required