CBSE 2024 · Region 2 · Set 1 · Q17 · 2 marks
A $\displaystyle 6$% solution of glucose (molar $\displaystyle \operatorname{mass}=180 \mathrm{~g} \mathrm{~mol}^{-1}$ ) is isotonic with $\displaystyle 2.5$% solution of an unknown organic substance. Calculate the molecular weight of the unknown organic substance.
Marking-scheme solution
\(\displaystyle \pi_{\text {Glucose }}=\pi_{\text {Unknown }}\)
\[\begin{array}{ll}
\mathrm{C}_{\mathrm{G}}=\mathrm{C}_{\mathrm{U}} & \frac{W_{\mathrm{G}}}{\mathrm{M}_{\mathrm{G}}}=\frac{W_{\mathrm{U}}}{\mathrm{M}_{\mathrm{U}}} \\
\frac{6}{180}=\frac{2 \cdot 5}{\mathrm{M}_{\mathrm{U}}} & \\
\mathrm{M}_{\mathrm{U}=} \frac{2 \cdot 5 \times 180}{6} \mathrm{~g} \mathrm{~mol}^{-1} & =75 \mathrm{~g} \mathrm{~mol}^{-1}
\end{array}
\]
SolutionsColligative Properties and Determination of Molar MassApplyvery_short_answermedium
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CBSE Class 12 Chemistry past-paper question from the 2024board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.