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Mole fraction of $I_2$ in $C_6H_6$ is 0.2.The molality of $I_2$ in $C_6H_6$ is

$\begin{array}{1 1}(a)\;3.2&(b)\;6.4\\(c)\;1.6&(d)\;2.3\end{array}$

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$\large\frac{n}{n+N}=$$0.2$
$\large\frac{N}{n+N}$$=0.8$
Or $\large\frac{n}{N}=\frac{1}{4}$
$\large\frac{nI_2}{WC_6H_6}$$\times MC_6H_6=\large\frac{1}{4}$
$\therefore$Molality =$\large\frac{nI_2}{wC_6H_6}$$\times 1000$
$\qquad\qquad=\large\frac{1}{4}\times\frac{1000}{MC_6H_6}$
Molality =$\large\frac{1}{4}\times \frac{1000}{78}$
$\qquad\;\;\;\;=3.2$
Hence (a) is the correct answer.
answered Oct 29, 2013 by sreemathi.v
 

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