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BITSAT 2021
Exam
BITSAT
Year
2021
Subject
Chemistry
Chapter
Chemical Kinetics

BITSAT 2021 · Chemistry · Chemical Kinetics

XmL of H_ 2 gas effuses through a hole in a container in 5 seconds....

XmL\mathrm{XmL} of H2\mathrm{H}_{2} gas effuses through a hole in a container in 5 seconds. The time taken for the effusion of the same volume of the gas specified below under identical conditions is :

Options

  1. A10 seconds : He
  2. B20 seconds: O2\mathrm{O}_{2}
  3. C25 seconds: CO
  4. D55 seconds : CO2\mathrm{CO}_{2}

Correct answer

B. 20 seconds: O2\mathrm{O}_{2}

Step-by-step solution

Under identical conditions, r1r2=M2M1\frac{r_{1}}{r_{2}}=\sqrt{\frac{\mathrm{M}_{2}}{\mathrm{M}_{1}}} As rate of diffusion is also inversely proportional to time, we will have, t2t1=M2‾M1\frac{t_{2}}{t_{1}}=\sqrt{\frac{\overline{\mathrm{M}_{2}}}{\mathrm{M}_{1}}} (A) Thus, For He,t2=42(5s)=52s≠10s\mathrm{He}, t_{2}=\sqrt{\frac{4}{2}}(5 s)=5 \sqrt{2} s \neq 10 s; (B) ForO2,t2=322(5s)=20s\mathrm{For} \mathrm{O}_{2}, t_{2}=\sqrt{\frac{32}{2}}(5 s)=20 s (C) For CO,t2=282(5s)≠25s\mathrm{CO}, t_{2}=\sqrt{\frac{28}{2}}(5 s) \neq 25 s; (D) For CO2,t2=442(5s)≠55s\mathrm{CO}_{2}, t_{2}=\sqrt{\frac{44}{2}}(5 s) \neq 55 s #
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