Question

Under identical conditions gaseous CO2 and CCl4 are allowed to effuse through a pinhole. If the rate of effusion of the CO2 is 6.3x10-2 mol/s, what is the rate of effusion of the CCl4?

Answer 1

According to Graham's law,

$\frac{R_1}{R_2}=\sqrt{\frac{M_2}{M_1}}$ ,

where R are rates of effusion, and M are molar masses of the corresponding gases.

Therefore,

$\frac{R_{CO_2}}{R_{CCl_4}}=\sqrt{\frac{M_{CCl_4}}{M_{CO_2}}}$

$\frac{6.3\cdot10^{-2}mol/s}{R_{CCl_4}}=\sqrt{\frac{153.82g/mol}{44.01g/mol}}$

$R_{CCl_4}=3.4\cdot10^{-2}mol/s$

Answer: $3.4\cdot10^{-2}mol/s$