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Q. 5% solution of sugar cane (mol. wt = 342) is isotonic with 1 % solution of X under similar conditions. The mol. mass of X is

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Solution:

$\pi_{1} = \pi_{2}$
$C_{1} = C_{2}
\frac{5 / 342}{0.1} = \frac{1/M}{0.1}$
$\frac{5}{342} = \frac{1}{M} \quad\Rightarrow \quad M = \frac{342}{5} = 68.4 \,gm/mol$