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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?

$(a)\;34.92\qquad(b)\;33.90\qquad(c)\;32\qquad(d)\;46.6$

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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 $O_2$
$\large\frac{r_{O_2}}{r_m}=\sqrt{(\large\frac{M_m}{M_{O_2}})}$
(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
 

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