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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Solutions
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A 2% solution of non electrolyte X is isotonic with 1% of urea(mol.wt.60),the molecular weight of X is

$\begin{array}{1 1}(a)\;240&(b)\;120\\(c)\;180&(d)\;60\end{array}$

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Answer:$120\; gm$
Two solutions having same osmotic pressure are consider isotonic. $\pi_1 = \pi_2$ (at same temperature)
If $\pi_1 = \pi_2 \rightarrow C_1RT = C_2RT \rightarrow C_1 = C_2 \rightarrow n_1 = n_2$ where n is the number of moles of solute, $n = \large\frac{w}{M}$
$\Rightarrow \large\frac{2 \%}{M_X} $$ = \large\frac{1\%}{60}$$ \rightarrow M_X = 120\; gm$
answered Feb 13, 2014 by sreemathi.v
edited Jul 16, 2014 by balaji.thirumalai
 

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