11. one mole of h2s gas escapes from a container by effusion in 77 seconds. how long would it take one mole…

Explanation:

Step1: Determine molar masses

The molar mass of $H_2S$, $M_{H_2S}=(2\times1 + 32)\text{ g/mol}=34\text{ g/mol}$. The molar mass of $NH_3$, $M_{NH_3}=(14 + 3\times1)\text{ g/mol}=17\text{ g/mol}$.

Step2: Apply Graham's law of effusion

Graham's law is $\frac{r_1}{r_2}=\sqrt{\frac{M_2}{M_1}}$, where $r$ is the rate of effusion and $M$ is the molar - mass. Since rate $r=\frac{n}{t}$ (where $n$ is the number of moles and $t$ is time), for the same number of moles ($n_1 = n_2 = 1$ mole), we can rewrite the law as $\frac{t_2}{t_1}=\sqrt{\frac{M_2}{M_1}}$. Let gas 1 be $H_2S$ and gas 2 be $NH_3$. We know $t_1 = 77$ s, $M_1 = 34\text{ g/mol}$, and $M_2 = 17\text{ g/mol}$. Then $t_2=t_1\sqrt{\frac{M_{NH_3}}{M_{H_2S}}}$.

Step3: Calculate the time for $NH_3$ to effuse

Substitute the values into the formula: $t_2 = 77\ s\times\sqrt{\frac{17\text{ g/mol}}{34\text{ g/mol}}}=77\ s\times\sqrt{\frac{1}{2}}\approx77\ s\times0.707 = 54.4\ s$.

Answer:

$54.4$ s