The volumes of ozone and chlorine gas diffusi | vedclass

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(B) According to Graham's Law of Diffusion,the rate of diffusion $r$ is proportional to the volume $V$ diffused in time $t$ $(r = V/t)$.Since the time

Vedclass · Accepted Answer

$48$

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(B) According to Graham's Law of Diffusion,the rate of diffusion $r$ is proportional to the volume $V$ diffused in time $t$ $(r = V/t)$.Since the time $t$ is the same for both gases,the ratio of rates is equal to the ratio of volumes: $\frac{r_1}{r_2} = \frac{V_1}{V_2}$.Also,$\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$,where $M$ is the molar mass.Equating the two,we get $\frac{V_{O_3}}{V_{Cl_2}} = \sqrt{\frac{M_{Cl_2}}{M_{O_3}}}$.Substituting the given values: $\frac{35}{29} = \sqrt{\frac{71}{M_{O_3}}}$.Squaring both sides: $(\frac{35}{29})^2 = \frac{71}{M_{O_3}}$.$M_{O_3} = 71 	imes (\frac{29}{35})^2$.$M_{O_3} = 71 	imes (0.82857)^2 \approx 71 	imes 0.6865 \approx 48.74$.Thus,the molar mass of ozone is approximately $48.7$.

The volumes of ozone and chlorine gas diffusing in the same time are $35 \ mL$ and $29 \ mL$ respectively. If the molar mass of chlorine is $71$,calculate the molar mass of ozone. (in $.7$)

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