The volumes of ozone and chlorine gas diffusi | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

$48$

Vedclass · Answer

(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 $.7$)

Similar Questions

At what temperature $(^oC)$ will the average velocity of oxygen molecules be equal to their $(rms)$ velocity at $27^oC$?

At what temperature $K$ will the $rms$ speed of $SO_2$ gas be equal to the $rms$ speed of $O_2$ gas at $303 \ K$?

What is the kinetic energy (in $J \ mol^{-1}$) of one mole of an ideal gas (molar mass $= 0.1 \ kg \ mol^{-1}$) if its rms velocity is $4 \times 10^2 \ ms^{-1}$ at $T(K)$?

Explain the following terms:
$(i) u_{mp}$
$(ii) u_{av}$
$(iii) \overline{u}^2$
$(iv) u_{rms}$

The ratio of the most probable speed,average speed,and $RMS$ speed is given by:

Vedclass Test Series

Exam Paper Generator

Online Exam Module