CBSE 2025 · Region 5 · Set 1 · Q17 · 2 marks
Calculate the elevation of boiling point of a solution when $\displaystyle 3$ g of $\displaystyle \mathrm{CaCl}_{2}\left(\right.$ Molar mass $\displaystyle \left.=111 \mathrm{~g} \mathrm{~mol}^{-1}\right)$ was dissolved in $\displaystyle 260$ g of water, assuming that $\displaystyle \mathrm{CaCl}_{2}$ undergoes complete dissociation. ( $\displaystyle \mathrm{K}_{\mathrm{b}}$ for water $\displaystyle =0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ )Liquids 'X' and 'Y' form an ideal solution. The vapour pressure of pure 'X' and pure 'Y' are $\displaystyle 120$ mm Hg and $\displaystyle 160$ mm Hg respectively. Calculate the vapour pressure of the solution containing equal moles of 'X' and 'Y'.
Calculate the elevation of boiling point of a solution when $\displaystyle 3$ g of $\displaystyle \mathrm{CaCl}_{2}\left(\right.$ Molar mass $\displaystyle \left.=111 \mathrm{~g} \mathrm{~mol}^{-1}\right)$ was dissolved in $\displaystyle 260$ g of water, assuming that $\displaystyle \mathrm{CaCl}_{2}$ undergoes complete dissociation. ( $\displaystyle \mathrm{K}_{\mathrm{b}}$ for water $\displaystyle =0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ )
Liquids 'X' and 'Y' form an ideal solution. The vapour pressure of pure 'X' and pure 'Y' are $\displaystyle 120$ mm Hg and $\displaystyle 160$ mm Hg respectively. Calculate the vapour pressure of the solution containing equal moles of 'X' and 'Y'.
Marking-scheme solution
ΔTb = iKb m
i=$\displaystyle 3$
Δ Tb =$\displaystyle 3$× $\displaystyle 0$·$\displaystyle 52$ ×3x $\displaystyle 1000$
=0.162K
Given nX = nY
χX = χy=$\displaystyle 0.5$
T
P = pX 𝑥
$\displaystyle 0$ 𝑥+pY 𝑥
$\displaystyle 0$ 𝑌 /
= $\displaystyle 120$ x $\displaystyle 0.5$ +$\displaystyle 160$ x $\displaystyle 0.5$
=140mm Hg
SolutionsColligative Properties and Determination of Molar MassApplyvery_short_answermedium
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CBSE Class 12 Chemistry past-paper question from the 2025board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.