Chapter 14, Problem 71A

Interpretation Introduction

Interpretation: The gas that effuses faster and how much faster between hydrogen and chlorine is to be determined.

Concept introduction: According to Graham's law, the gas diffusing rate, or effusing rate, at a given temperature is inversely related to the square root of the gas's molecular mass.

Answer to Problem 71A

The hydrogen gas effuses faster than chlorine. The effusion of hydrogen gas is $5.9582$ times faster than chlorine gas.

Explanation of Solution

The molecular mass of hydrogen gas is $2{\text{ g mol}}^{-1}$ .

The molecular mass of chlorine gas is ${\text{71 g mol}}^{-1}$ .

Since hydrogen gas has a smaller molecular mass than chlorine gas, it effuses faster than chlorine gas.

The formula to calculate the rate of effusion is as follows:

  $\frac{Rat{e}_{{\text{H}}_{\text{2}}}}{Rat{e}_{{\text{Cl}}_{\text{2}}}}=\sqrt{\frac{M_{{\text{Cl}}_{\text{2}}}}{M_{{\text{H}}_{\text{2}}}}}$

Where,

• $M_{{\text{H}}_{\text{2}}}$ is the molecular mass of hydrogen gas.
• $M_{{\text{Cl}}_{\text{2}}}$ is the molecular mass of chlorine gas.
• $Rat{e}_{{\text{H}}_{\text{2}}}$ is the rate of effusion of hydrogen gas.
• $Rat{e}_{{\text{Cl}}_{\text{2}}}$ is the rate of effusion of chlorine gas.

Substitute the values of $M_{{\text{H}}_{\text{2}}}$ and $M_{{\text{Cl}}_{\text{2}}}$ in the above equation as follows:

  $\begin{array}{c}\frac{Rat{e}_{{\text{H}}_{\text{2}}}}{Rat{e}_{{\text{Cl}}_{\text{2}}}}=\sqrt{\frac{M_{{\text{Cl}}_{\text{2}}}}{M_{{\text{H}}_{\text{2}}}}}\\ =\sqrt{\frac{\text{71 }\bcancel{{\text{g mol}}^{-1}}}{\text{2 }\bcancel{{\text{g mol}}^{-1}}}}\\ =\sqrt{35.5}\\ =5.9582\end{array}$

Therefore, hydrogen gas effuses $5.9582$ times faster than chlorine gas.

Conclusion

The hydrogen gas effuses faster than chlorine. The effusion of hydrogen gas is $5.9582$ times faster than chlorine gas.