A $25 \%$ solution of cane-sugar (molar mass $=342 \mathrm{~g} \mathrm{~mol}^{-1}$ ) is isotonic with $5 \%$…
A $25 \%$ solution of cane-sugar (molar mass $=342 \mathrm{~g} \mathrm{~mol}^{-1}$ ) is isotonic with $5 \%$ solution of a substance $A$. Then find the molecular weight of $A$.
$6.84 \mathrm{~g} \mathrm{~mol}^{-1}$
$68.4 \mathrm{~g} \mathrm{~mol}^{-1}$
$25 \mathrm{~g} \mathrm{~mol}^{-1}$
$684 \mathrm{~g} \mathrm{~mol}^{-1}$
Solution
Two solutions are isotonic when they have the same osmotic pressure. It is only possible when they have same molar concentration.
A $25 \%$ solution of cane-sugar means $100 \mathrm{~g}$ of solution contain $25 \mathrm{~g}$ of cane-sugar.
For dilute solution $100 \mathrm{~g}$ is approximately equal to $100 \mathrm{~mL}$.
Molarity of cane-sugar
$
\begin{aligned}
& =\frac{\text { Number of mole of cane-sugar }}{\text { Volume (in L) }} \\
& =\frac{25 \mathrm{~g}}{342 \mathrm{~g} / \mathrm{mol} \times 100 \mathrm{~mL}} \times 1000 \mathrm{~mL} \\
& =0.7309 \mathrm{M}...(i)
\end{aligned}
$
Now, $5 \%$ solution of substance $A$ means $100 \mathrm{~g}$ of the solution contain $5 \mathrm{~g}$ of substance $A$.
For dilute solution, $100 \mathrm{~g}$ is approximately equal to $100 \mathrm{~mL}$
Molarity o
$
\text { of } \begin{aligned}
A & =\frac{5 \mathrm{~g}}{M^{\prime} \mathrm{g} / \mathrm{mol} \times 100 \mathrm{~mL}} \times 1000 \mathrm{~mL} \\
& =\frac{50}{M^{\prime}} \mathrm{M}...(ii)
\end{aligned}
$
Equating both equations (i) and (ii), we get
$
\begin{aligned}
\frac{50}{M^{\prime}} & =0.7309 \\
M^{\prime} & =\frac{50}{0.7309} \\
& =68.408 \mathrm{~g} \mathrm{~mol}^{-1}
\end{aligned}
$
Hence, molecular weight of $A$ is $68.4 \mathrm{~g} \mathrm{~mol}^{-1}$