Question

Using Graham's Law of Effusion, calculate
the approximate time it would take for
1.0 L of argon gas to effuse, if 1.0 L of
oxygen gas took 12.7 minutes to effuse
through the same opening.

0.070 minutes

0.89 minutes

None of the other answers

14 minutes

12 minutes

Answer 1

The rate of effusion of Argon here is 8.03 min

Data;

• Molar mass of Oxygen = 16g/mol
• Time for effusion of oxygen = 12.7 min
• Molar mass of Argon = 40g/mol
• Time for effusion of Argon = ?

Graham's Law of Effusion

This law states that the rate of effusion of a gas is inversely proportional to the square root of it's molar mass.

$R \alpha \frac{1}{\sqrt{M} }$

From this,

$\frac{r_1}{r_2} = \frac{\sqrt{M_2} }{M_1}$

substituting the values and solving,

$\frac{R_A_r}{12.7}= \frac{\sqrt{16} }{\sqrt{40} } \\ R_A_r = \frac{12.7*\sqrt{16} }{\sqrt{40} } \\R_A_r = 8.032min$

The rate of effusion of Argon here is 8.03 min

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