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A solution containing 10 g per \(\mathrm{dm}^3\) of urea (molecular mass \(=60 \mathrm{~g} \mathrm{~mol}^{-1}\)) is isotonic with a \(5 \%\) solution of a nonvolatile solute. The molecular mass of this nonvolatile solution is:
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Verified Answer
The correct answer is:
\(350 \mathrm{~g} \mathrm{~mol}^{-1}\)
The molar concentration of urea solution
$$
=\frac{10 g / d m^3}{\text { Mol. wt. of urea }}=\frac{10}{60} \mathrm{M}=\frac{1}{6} \mathrm{M}
$$
Molar concentration of $5 \%$ nonvolatile solute
$$
\begin{aligned}
& =\frac{50 \mathrm{~g} / \mathrm{dm}^3}{\text { mol. wt. of non - volatile solute }} \\
& =\frac{50}{\mathrm{~m}} \mathrm{M}
\end{aligned}
$$
Both solutions are isotonic to each other;
therefore, $\quad \frac{1}{6}=\frac{50}{m}$
$$
m=50 \times 6=300 \mathrm{~g} \mathrm{~mol}^{-1}
$$
Caution
$10 \mathrm{~g}$ per $\mathrm{dm}^3$ of urea is isotonic with $5 \%$ solution of a non-volatile solute. Hence, osmosis cannot occur between the two solutions. Thus, their molar concentrations are equal to each other.
$$
=\frac{10 g / d m^3}{\text { Mol. wt. of urea }}=\frac{10}{60} \mathrm{M}=\frac{1}{6} \mathrm{M}
$$
Molar concentration of $5 \%$ nonvolatile solute
$$
\begin{aligned}
& =\frac{50 \mathrm{~g} / \mathrm{dm}^3}{\text { mol. wt. of non - volatile solute }} \\
& =\frac{50}{\mathrm{~m}} \mathrm{M}
\end{aligned}
$$
Both solutions are isotonic to each other;
therefore, $\quad \frac{1}{6}=\frac{50}{m}$
$$
m=50 \times 6=300 \mathrm{~g} \mathrm{~mol}^{-1}
$$
Caution
$10 \mathrm{~g}$ per $\mathrm{dm}^3$ of urea is isotonic with $5 \%$ solution of a non-volatile solute. Hence, osmosis cannot occur between the two solutions. Thus, their molar concentrations are equal to each other.
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