# Question 8c69f

Oct 13, 2015

$\text{36 g/mol}$

#### Explanation:

The idea here is that the rate of effusion of a gas, that is, the number of moles of gas per unit of time, is inversely proportional to the square root of its molar mass - this is known as Graham's Law.

$\text{rate} \propto \frac{1}{\sqrt{{M}_{M}}}$

In other words, the heavier each individual molecule of a gas is, the slower it will effuse. LIkewise, the lighter the molecules of a gas, the faster they will effuse.

Here's how that would look for helium and ethylene oxide - notice that fewer molecules of ethylene oxide effuse when compared with the lighter helium molecules.

In your case, you know that a certain gas takes three times as long to effuse as jelium. Right from the start, you know that you're dealing with a gas that has heavier molecules than helium does.

If you take ${r}_{H e}$ to be the rate of effusion of helium, and ${r}_{X}$ to be the rate of effusion of the unknown gas, you can write

${r}_{H e} = 3 \times {r}_{X}$

Moreover, you know that

r_(He) prop 1/sqrt(M_"M helium")" " and " "r_(X) prop 1/sqrt(M_"M gas X")

If you divide these expressions, you will find that

${r}_{H e} / {r}_{X} = \frac{1}{\sqrt{{M}_{\text{M helium") * sqrt(M_"M gas X}}}}$

$\frac{3 \times \textcolor{red}{\cancel{\textcolor{b l a c k}{{r}_{X}}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{{r}_{X}}}}} = \sqrt{{M}_{\text{M gas X")/sqrt(M_"M helium}}}$

To keep the calculations simple, you can use helium's molar mass to be $\text{4 g/mol}$.

Rearrange the above equation to solve for ${M}_{\text{M gas X}}$

$\sqrt{{M}_{\text{M gas X") = 3 xx sqrt(M_"M helium}}}$

Square both sides to get rid of the square roots

${\left(\sqrt{{M}_{\text{M gas X"))^2 = 3^2 xx (sqrt(M_"M helium}}}\right)}^{2}$

M_"M gas X" = 9 xx "4 g/mol" = color(green)("36 g/mol")#