Effusion - SAT Subject Test in Chemistry

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Question

How much faster is the rate of effusion of $H_2 _{(g)}$ than the rate of effusion of $H_2O_{(g)}$ ?

Answer

By Graham's Law, $\frac{Rate_1}{Rate_2}=\sqrt{\frac{M_2}{M_1}}$ . The molar mass of $H_2_{(g)}$ is 2 g/mol and the molar mass of $H_2O_{(g)}$ is 18 g/mol. Thus, $\sqrt{\frac{18}{2}}=\sqrt{9}=3$

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Question

How much faster is the rate of effusion of $H_2 _{(g)}$ than the rate of effusion of $H_2O_{(g)}$ ?

Answer

By Graham's Law, $\frac{Rate_1}{Rate_2}=\sqrt{\frac{M_2}{M_1}}$ . The molar mass of $H_2_{(g)}$ is 2 g/mol and the molar mass of $H_2O_{(g)}$ is 18 g/mol. Thus, $\sqrt{\frac{18}{2}}=\sqrt{9}=3$

Compare your answer with the correct one above

Question

How much faster is the rate of effusion of $H_2 _{(g)}$ than the rate of effusion of $H_2O_{(g)}$ ?

Answer

By Graham's Law, $\frac{Rate_1}{Rate_2}=\sqrt{\frac{M_2}{M_1}}$ . The molar mass of $H_2_{(g)}$ is 2 g/mol and the molar mass of $H_2O_{(g)}$ is 18 g/mol. Thus, $\sqrt{\frac{18}{2}}=\sqrt{9}=3$

Compare your answer with the correct one above

Question

How much faster is the rate of effusion of $H_2 _{(g)}$ than the rate of effusion of $H_2O_{(g)}$ ?

Answer

By Graham's Law, $\frac{Rate_1}{Rate_2}=\sqrt{\frac{M_2}{M_1}}$ . The molar mass of $H_2_{(g)}$ is 2 g/mol and the molar mass of $H_2O_{(g)}$ is 18 g/mol. Thus, $\sqrt{\frac{18}{2}}=\sqrt{9}=3$

Compare your answer with the correct one above

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