We’re being asked to calculate **molar mass of gas Y** if it **effuse****s half as fast as**** O_{2}**.

Recall that ** Graham's Law of Effusion** allows us to compare the rate of effusion of two gases. Graham's Law states that the rate of effusion of a gas is inversely proportional to its molar mass.

$\mathbf{rate}\mathbf{=}\frac{\mathbf{1}}{\sqrt{{\mathbf{MM}}_{\mathbf{gas}}}}$

This means that when comparing two gases:

$\overline{)\frac{{\mathbf{rate}}_{\mathbf{gas}\mathbf{}\mathbf{1}}}{{\mathbf{rate}}_{\mathbf{gas}\mathbf{}\mathbf{2}}}{\mathbf{=}}\sqrt{\frac{{\mathbf{MM}}_{\mathbf{gas}\mathbf{}\mathbf{2}}}{{\mathbf{MM}}_{\mathbf{gas}\mathbf{}\mathbf{1}}}}}$

Let's designate O_{2} as gas 2 and the gas Y as gas 1.

**rate gas Y = 0.5 (rate O _{2})**

- $\frac{{\mathbf{rate}}_{\mathbf{gas}\mathbf{}\mathbf{1}}}{{\mathbf{rate}}_{\mathbf{gas}\mathbf{}\mathbf{2}}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{5}$

The molar mass of O_{2} is:

O_{2}**2** O × 16 g/mol O = __32 g/mol __

** Sum = ****32 g/mol**

Gas Y effuses half as fast as O2. What is the molar mass of gas Y?

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Effusion concept. You can view video lessons to learn Effusion. Or if you need more Effusion practice, you can also practice Effusion practice problems.