3.2 g of oxygen is diffused in 10 min. Unde | vedclass

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(A) According to Graham's Law of diffusion,the rate of diffusion $r$ is given by $r = \frac{w}{M 	imes t}$,where $w$ is the mass,$M$ is the molar mas

Vedclass · Accepted Answer

$9.3$

Vedclass · Answer

(A) According to Graham's Law of diffusion,the rate of diffusion $r$ is given by $r = \frac{w}{M 	imes t}$,where $w$ is the mass,$M$ is the molar mass,and $t$ is the time.For the same volume of gas,the ratio of rates is $\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$.Substituting the values: $\frac{w_1 / (M_1 	imes t_1)}{w_2 / (M_2 	imes t_2)} = \sqrt{\frac{M_2}{M_1}}$.Given: $w_{O_2} = 3.2 \ g$,$M_{O_2} = 32 \ g/mol$,$t_{O_2} = 10 \ min$,$w_{N_2} = 2.8 \ g$,$M_{N_2} = 28 \ g/mol$.$\frac{3.2 / (32 	imes 10)}{2.8 / (28 	imes t_{N_2})} = \sqrt{\frac{28}{32}}$.$\frac{0.1 / 10}{0.1 / t_{N_2}} = \sqrt{0.875}$.$\frac{t_{N_2}}{10} = 0.9354$.$t_{N_2} = 9.35 \ min \approx 9.3 \ min$.

$3.2 \ g$ of oxygen is diffused in $10 \ min$. Under similar conditions,$2.8 \ g$ of nitrogen will diffuse in .............. $min$.

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