Question 21958

May 2, 2016

${\text{100 g mol}}^{- 1}$

Explanation:

The idea here is that you need to use Graham's Law of Diffusion to find a relationship between the volumes of the two gases and their molar masses.

When two gases are kept under the same conditions for pressure and temperature, the ratio that exists between their number of moles is equivalent to the ratio that exists between their volumes.

You can prove this by using the ideal gas law equation

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} P V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Here

$P$ - the pressure of the gas
$V$ - the volume it occupies
$n$ - the number of moles of gas
$R$ - the universal gas constant, usually given as $0.0821 \left(\text{atm" * "L")/("mol" * "K}\right)$
$T$ - the absolute temperature of the gas

You can write

$P \cdot {V}_{A} = {n}_{A} \cdot R T \to$ for gas $\text{A}$

$P \cdot {V}_{B} = {n}_{B} \cdot R T \to$ for gas $\text{B}$

Divide these two equations to get

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{P}}} \cdot {V}_{A}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{P}}} \cdot {V}_{B}} = \frac{{n}_{A} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{R T}}}}{{n}_{B} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{R T}}}} \implies \textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{V}_{A} / {V}_{B} = {n}_{A} / {n}_{B}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

This means that the diffusion of a volume of gas is equivalent to the diffusion of a number of moles of gas.

Now, according to Graham's Law of Diffusion, the rate of diffusion of a gas is inversely proportional to the square root of its density

color(blue)(|bar(ul(color(white)(a/a)"rate" prop 1/sqrt("density")color(white)(a/a)|)))

You can use the ideal gas law equation to write the density of a gas as a function of its molar mass. Keep in mind that you have

$n = \frac{m}{M} _ M$

Here $m$ is mass and ${M}_{M}$ is molar mass. You will thus have

$P V = \frac{m}{M} _ M R T$

Rearrange to get

$P \cdot {M}_{M} = \frac{m}{V} \cdot R T$

$\frac{P}{R T} \cdot {M}_{M} = {\overbrace{\frac{m}{V}}}^{\textcolor{b l u e}{\text{density}}}$

So, the rate of diffusion can be expressed in terms of the volumes of the two gases and of their molar masses

$\text{rate"_A prop 1/sqrt(P/(RT) * M_"M A")" }$ and " " "rate"_B prop 1/sqrt(P/(RT) * M_"M B")

You can say that if $\text{50 mL}$ of gas $\text{A}$ diffuses in the same period of time as $\text{40 mL}$ of gas $\text{B}$, the ratio of their rates of diffusion will be

"rate"_A/"rate"_B = (50 color(red)(cancel(color(black)("mL"))))/(40color(red)(cancel(color(black)("mL")))) = 5/4

This is equivalent to

$\frac{5}{4} = \sqrt{\frac{P}{R T} \cdot {M}_{\text{M B")/sqrt(P/(RT) * M_"M A") = color(red)(cancel(color(black)(sqrt(P/(RT)))))/color(red)(cancel(color(black)(sqrt(P/(RT))))) * sqrt(M_"M B")/sqrt(M_"M A}}}$

Since you know that

${M}_{\text{M A" = "64 g mol}}^{- 1}$

you can say that

$\frac{5}{4} = \frac{\sqrt{{M}_{\text{M B}}}}{\sqrt{64}}$

This will get you

$\sqrt{{M}_{\text{M B}}} = \frac{5}{4} \cdot \sqrt{64} = 10$

Therefore,

M_"M B" = 10^2 = color(green)(|bar(ul(color(white)(a/a)"100 g mol"^(-1)color(white)(a/a)|)))#