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Chemistry · 2024 · 2 marks
CBSE 2024 · Region 2 · Set 3 · Q21
A 3%solution of glucose (molar mass =180 g mol^-1 ) is isotonic with 2.5% solution of an unknown organic substance. Calculate the molecular weight of…
A $\displaystyle 3$%solution of glucose (molar mass $\displaystyle =180 \mathrm{~g} \mathrm{~mol}^{-1}$ ) is isotonic with $\displaystyle 2.5$% solution of an unknown organic substance. Calculate the molecular weight of the unknown organic substance.
Marking-scheme solution
\[\begin{aligned}
& \pi_{1}=\pi_{2} \\
& \frac{W_{1}}{M_{1}}=\frac{W_{2}}{M_{2}} \\
& \frac{3}{180}=\frac{2 \cdot 5}{M_{2}} \\
& M_{2}=\frac{2 \cdot 5 \times 180}{3}=150 \mathrm{~g} \mathrm{~mol}^{-1}
\end{aligned}
\]
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CBSE Class 12 Chemistry past-paper question from the 2024 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.