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Chemistry Β· 2026 Β· 3 marks
CBSE 2026 Β· Region 2 Β· Set 1 Β· Q22
When 1.5 g of a non-volatile solute was dissolved in 90 g of benzene, the boiling point of benzene got raised from 353.23 K to 353.93 K. Calculateβ¦
When $\displaystyle 1.5$ g of a non-volatile solute was dissolved in $\displaystyle 90$ g of benzene, the boiling point of benzene got raised from $\displaystyle 353.23$ K to $\displaystyle 353.93$ K. Calculate the molar mass of the solute. (Given $\displaystyle \mathrm{K}_{\mathrm{b}}$ for benzene $\displaystyle =2.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-\mathbf{1}}$ )
Marking-scheme solution
π₯ Tπ= Kπ π ππ΅
ππ΅
π $\displaystyle 1000$
ππ΄
π₯ Tπ= ππ β π$\displaystyle 0$π =$\displaystyle 353.93$-$\displaystyle 353.23$=$\displaystyle 0.70$ K
$\displaystyle 0.70$ = $\displaystyle 2.52$ X $\displaystyle 1.5$
ππ΅
π $\displaystyle 1000$
ππ΅= $\displaystyle 2.52$ X $\displaystyle 1.5$ X $\displaystyle 1000$
$\displaystyle 0.70$ X $\displaystyle 90$
ππ΅= $\displaystyle 3780$
MB = $\displaystyle 60$ g molβ$\displaystyle 1$
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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.