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Chemistry · 2024 · 2 marks
CBSE 2024 · Region 2 · Set 2 · Q21
A 4% solution of urea (molar mass =60 g mol^-1 ) is isotonic with 5.2% solution of an unknown organic substance. Calculate the molecular weight of…
A $\displaystyle 4$% solution of urea (molar mass $\displaystyle =60 \mathrm{~g} \mathrm{~mol}^{-1}$ ) is isotonic with $\displaystyle 5.2$% solution of an unknown organic substance. Calculate the molecular weight of the unknown organic substance.
Marking-scheme solution
\[\begin{aligned}
& \pi_{1}=\pi_{2} \\
& \frac{w_{\text {urea }}}{M_{\text {urea }}}=\frac{w_{A}}{M_{A}} \\
& \mathbf{M}_{\mathbf{A}}=\mathbf{5 . 2} \times \mathbf{1 5}=\mathbf{7 8 . 0 ~ \mathbf { g ~ m o l } ^ { - \mathbf { 1 } }}
\end{aligned}
\]
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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.