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Chemistry · 2025 · 3 marks
CBSE 2025 · Region 1 · Set 1 · Q22
A solution of glucose (molar mass =180 g mol^-1 ) in water has a boiling point of 100.20^° C. Calculate the freezing point of the same solution.…
A solution of glucose (molar mass $\displaystyle =180 \mathrm{~g} \mathrm{~mol}^{-1}$ ) in water has a boiling point of $\displaystyle 100.20^{\circ} \mathrm{C}$. Calculate the freezing point of the same solution. Molal constants for water $\displaystyle \mathrm{K}_{\mathrm{f}}$ and $\displaystyle \mathrm{K}_{\mathrm{b}}$ are $\displaystyle 1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ and $\displaystyle 0.512 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ respectively.
Marking-scheme solution
Tb of glucose solution = $\displaystyle 100.20$°C
Δ Tb =Tb - Tob
= $\displaystyle 100.20$ o – $\displaystyle 100$ oC = $\displaystyle 0.20$ oC or $\displaystyle 0.20$ K
△Tb = Kb .m
m = $\displaystyle 0.20$ = $\displaystyle 0.390$ mol/kg
Or -$\displaystyle 0.725$ oC
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CBSE Class 12 Chemistry past-paper question from the 2025 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.