# One of two identical balloons contained carbon dioxide (CO2, 44 g mol−1) and the other contained hydrogen (H2, 2.0 g mol−1). If it took 24 hours for all of the H2 to escape from its balloon, how long did it take for all of the CO2 to escape?

Apr 21, 2017

$112.569978 h o u r s$ which rounds to $110 h o u r s$ due to significant figures.

#### Explanation:

Use Graham's Law:
${r}_{1} / {r}_{2} = {\sqrt{M}}_{2} / {\sqrt{M}}_{1} = {t}_{2} / {t}_{1}$
Note: $r$ is the rate of effusion, $M$ is the molar mass of the gas, and $t$ is the time the gas took to effuse. Also, know that this formula only works if both gasses are at the same temperature.

Since molar mass and time is given, we will use the second half of the formula.

$\sqrt{\text{molar mass" CO_2)/sqrt("molar mass} {H}_{2}} = \frac{t i m e C {O}_{2}}{t i m e {H}_{2}}$

$\frac{\sqrt{44 g \cdot m o {l}^{- 1} C {O}_{2}}}{\sqrt{2.0 g \cdot m o {l}^{- 1} {H}_{2}}} = \frac{t i m e C {O}_{2}}{24 h o u r s {H}_{2}}$

$\frac{\left(24 h o u r s {H}_{2}\right) \cdot \sqrt{44 g \cdot m o {l}^{- 1} C {O}_{2}}}{\sqrt{2.0 g \cdot m o {l}^{- 1} {H}_{2}}} = t i m e C {O}_{2}$

$t i m e C {O}_{2} = 112.569978 h o u r s$

Since there are only 2 significant figures the answer would be $110 h o u r s$