Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

Pure $O_2$ diffuses through an aperture in 224 seconds whereas mixture of $O_2$ and another gas containing $80\%\;O_2$ diffuses from the same in 234 sec. What is mol. wt. of gas?


Can you answer this question?

1 Answer

0 votes
For gaseous mixture $80\% O_2 , 20\%$ gas.
$\therefore \text{average mol. wt of mixture}$
$(M_m) = \large\frac{32\times80+20\times m}{100} $----(i)
Now for diffusion of gaseous mixture and pure $O_2$
(or) $\large\frac{V_{O_2}}{t_{O_2}}\times \large\frac{t_m}{V_m} = \sqrt{(\large\frac{M_m}{M_{O_2}})}$
(Since same volume diffuses)
$\Rightarrow \large\frac{1}{224}\times\large\frac{234}{1} = \sqrt{\large\frac{M_m}{32}}$
$M_m = 34.92$----(ii)
From eq (i) and (ii) we get
Mol wt of gas m = 46.6
Hence answer is (D)
answered Apr 5, 2014 by sharmaaparna1

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App