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Chemistry Β· 2026 Β· 3 marks
CBSE 2026 Β· Region 2 Β· Set 3 Β· Q22
60 g of glucose is dissolved in 250 g of water. Calculate the freezing point of this solution. ( molar mass of glucose =180 g mol^-1, K_f for waterβ¦
$\displaystyle 60$ g of glucose is dissolved in $\displaystyle 250$ g of water. Calculate the freezing point of this solution. \[\left(\text { molar mass of glucose }=180 \mathrm{~g} \mathrm{~mol}^{-1}, \mathrm{~K}_{\mathrm{f}} \text { for water }=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right) \]
Marking-scheme solution
ΞTπ= πΎπ ππ΅
ππ΅
π $\displaystyle 1000$
ππ΄
ΞTπ= 1.86X $\displaystyle 60$
$\displaystyle 180$π $\displaystyle 1000$
=
= $\displaystyle 2.48$Ξ
β³ππ= $\displaystyle 2.48$πΎ
Freezing point of solution = Tf0-β³Tf
= $\displaystyle 273.15$ K β $\displaystyle 2.48$ K / $\displaystyle 273$ K β $\displaystyle 2.48$ K
= 270.67K / $\displaystyle 270.52$ K or -$\displaystyle 2.48$ 0C
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CBSE Class 12 Chemistry past-paper question from the 2026 board exam, with the answer as CBSEβs own marking scheme gives it. Where our answers come from.