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?

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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?

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Answer 1 · 2017-06-23T20:58:56

It could be iodine..

Explanation:

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

$\frac{r_{1}}{r_{2}} = \sqrt{\frac{M M_{2}}{M M_{1}}}$ $(r_1)/(r_2) = sqrt((MM_2)/(MM_1))$

We can substitute gas $X$ $"X"$ and $oxygen$ $"oxygen"$ in for 1 and 2, respectively, to get

$\frac{r_{X}}{r_{O_{2}}} = \sqrt{\frac{M M_{O_{2}}}{M M_{X}}}$ $(r_"X")/(r_ ("O"_2)) = sqrt((MM_ ("O"_2))/(MM_"X"))$

Since $O_{2}$ $"O"_2$ effuses at a rate twice that of gas $X$ $"X"$ , the ratio of the rate of effusion of $X$ $"X"$ to $O_{2}$ $"O"_2$ is $0.5$ $0.5$ :

$0.5 = \sqrt{\frac{M M_{O_{2}}}{M M_{X}}}$ $0.5 = sqrt((MM_ ("O"_2))/(MM_"X"))$

Since the molar mass of $O_{2}$ $"O"_2$ is given as $32.00$ $32.00$ $g/mol$ $"g/mol"$ , we then have

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

Using algebra to solve for $M M_{X}$ $MM_"X"$ :

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

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

$M M_{X} = {(\frac{\sqrt{32 l g/mol}}{0.5})}^{2} = 128$ $MM_"X" = (sqrt(32color(white)(l)"g/mol")/0.5)^2 = color(red)(128$ $g/mol$ $color(red)("g/mol"$

The molar mass of the unknown gas is thus $128$ $128$ $g/mol$ $"g/mol"$

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

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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?

1 Answer

Explanation:

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