Answer to Question #132460 in General Chemistry for martha

Question

what is the molar mass of a gas that effuses through a small hole at twice the rate as oxygen gas O2 at the same temperature?

Answer 1

Molar mass of oxygen = 32g

According to Graham's Law of Effusion, the rate of effusion of a gas is inversely proportional to the square root of its molar mass.

$rate\alpha\dfrac{1}{\sqrt{\smash[b]{Mm}}}$

In our case; Mm O₂ =32g/mol

Mm gas X = ?

Now, since both gases are kept under the same conditions for pressure and temperature and they must effuse through the same hole, so you can say that their respective rates of effusion will relate as in the following equation;

$\dfrac{rateX}{rate O2} = \dfrac{\sqrt{\smash[b]{MmO2}}}{\sqrt{\smash[b]{MmX}}}$

Now from the question, we know that if rate of gas X is twice that of O₂

That means rate of X/rate of O₂ can be written as 2/1

$\dfrac{2}{1} = \dfrac{\sqrt{\smash[b]{32.0g}}}{\sqrt{\smash[b]{MmX}}}$ (Making $\sqrt{\smash[b]{MmX}}$ the subject gives;)

$\sqrt{\smash[b]{MmX}} =\dfrac{\sqrt{\smash[b]{32.0g}}}{2}$

$\sqrt{\smash[b]{MmX}} = 2.828g/mol$ (Removing the square root by squaring both sides)

$MmX =$ (2.828)²g/mol

MmX = 7.997g/mol

Molar mass of the unknown gas is approximately 8g/mole