$(a)\;40.4\times10^{-8}cm\qquad(b)\;41.4\times10^{-8}cm\qquad(c)\;42.4\times10^{-8}cm\qquad(d)\;44.4\times10^{-8}cm$

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

Since $6.023\times10^23$ molecules of $N_2$ occupy $22400cm^3$

One molecule of $N_2$ occupies $\large\frac{22400}{6.023\times10^23} cm^3$

(or) Volume of one molecule of $N_2 = 3.73\times 10^{-20}cm^3$

Also the average distance between two molecules = 2r

and $\large\frac{4}{3}\pi r^3 = 3.72\times10^{-20}$

$r^3 = \large\frac{3.72\times10^{-20}\times3\times7}{4\times22}$

$r = 20.7\times10^{-8}cm$

Thus average distance = $2\times20.7\times10^{-8}$

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\Rightarrow = 41.4\times10^{-8}cm$

Hence answer is (b)

Ask Question

Tag:MathPhyChemBioOther

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...