# In an experiment, an unknown gas effuses at one-half the speed of oxygen gas, which has a molar mass of 32 g/mol. What might the unknown gas be?

Jun 23, 2017

It could be iodine..

#### Explanation:

We can solve this equation using Graham's law of effusion:

$\frac{{r}_{1}}{{r}_{2}} = \sqrt{\frac{M {M}_{2}}{M {M}_{1}}}$

We can substitute gas $\text{X}$ and $\text{oxygen}$ in for 1 and 2, respectively, to get

(r_"X")/(r_ ("O"_2)) = sqrt((MM_ ("O"_2))/(MM_"X"))

Since ${\text{O}}_{2}$ effuses at a rate twice that of gas $\text{X}$, the ratio of the rate of effusion of $\text{X}$ to ${\text{O}}_{2}$ is $0.5$:

0.5 = sqrt((MM_ ("O"_2))/(MM_"X"))

Since the molar mass of ${\text{O}}_{2}$ is given as $32.00$ $\text{g/mol}$, we then have

$0.5 = \sqrt{\left(32 \textcolor{w h i t e}{l} \text{g/mol")/(MM_"X}\right)}$

Using algebra to solve for $M {M}_{\text{X}}$:

0.5 = sqrt(32color(white)(l)"g/mol")/(sqrt(MM_"X"))

$\frac{\sqrt{M {M}_{\text{X") = sqrt(32color(white)(l)"g/mol}}}}{0.5}$

MM_"X" = (sqrt(32color(white)(l)"g/mol")/0.5)^2 = color(red)(128 color(red)("g/mol"

The molar mass of the unknown gas is thus $128$ $\text{g/mol}$

A gas close to this value is sfcolor(blue)("iodine", with molar mass $126.90$ $\text{g/mol}$.