Test Index

BITSAT 2021 Paper

Section: Chemistry

© examsnet.com

Question : 74 of 150

Marks: +1, -0

X\,\mathrm{mL}

\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 :

10 seconds : He
20 seconds: $\mathrm{O}_2$
25 seconds: $\mathrm{CO}$
55 seconds : $\mathrm{CO}_2$

Solution:

Under identical conditions,

\frac{r_1}{r_2}=\sqrt{\frac{M_2}{M_1}}

As rate of diffusion is also inversely proportional to time, we will have,

\frac{t_2}{t_1}=\sqrt{\frac{M_2}{M_1}}

(a) Thus, For He,

t_2=\sqrt{\frac{4}{2}}(5\,\mathrm{s})=5\sqrt{2}\,\mathrm{s}\neq 10\,\mathrm{s}

;

(b) For

\mathrm{O}_2,\ t_2=\sqrt{\frac{32}{2}}(5\,\mathrm{s})=20\,\mathrm{s}

(c) For

\mathrm{CO},\ t_2=\sqrt{\frac{28}{2}}(5\,\mathrm{s})\neq 25\,\mathrm{s}

;

(d) For

\mathrm{CO}_2,\ t_2=\sqrt{\frac{44}{2}}(5\,\mathrm{s})\neq 55\,\mathrm{s}

© examsnet.com

Go to Question:

Prev Question

Next Question