CBSE 2026 · Region 5 · Set 1 · Q31 · 5 marks
(i)Calculate the freezing point of a solution when $\displaystyle 10 \cdot 5 \mathrm{~g}$ of $\displaystyle \mathrm{MgBr}_{2}$ was dissolved in $\displaystyle 250$ g of water, assuming $\displaystyle \mathrm{MgBr}_{2}$ undergoes complete dissociation. (Given : Molar mass of $\displaystyle \mathrm{MgBr}_{2}=184 \mathrm{~g} \mathrm{~mol}^{-1}$, $\displaystyle \mathrm{K}_{\mathrm{f}}$ for water $\displaystyle =1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ )(ii)Write two differences between ideal and non-ideal solutions.(i)A solution is prepared by dissolving $\displaystyle 0.025$ g of potassium sulphate in $\displaystyle 2$ L of water at $\displaystyle 27$°C. Assuming potassium sulphate is completely dissociated, determine its osmotic pressure. (Given : $\displaystyle \mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$, Molar mass of $\displaystyle \mathrm{K}_{2} \mathrm{SO}_{4}=174 \mathrm{~g} \mathrm{~mol}^{-1}$ )(ii)What type of azeotrope will be formed by a solution of acetone and chloroform ? Give reason. []
(i)
Calculate the freezing point of a solution when $\displaystyle 10 \cdot 5 \mathrm{~g}$ of $\displaystyle \mathrm{MgBr}_{2}$ was dissolved in $\displaystyle 250$ g of water, assuming $\displaystyle \mathrm{MgBr}_{2}$ undergoes complete dissociation. (Given : Molar mass of $\displaystyle \mathrm{MgBr}_{2}=184 \mathrm{~g} \mathrm{~mol}^{-1}$, $\displaystyle \mathrm{K}_{\mathrm{f}}$ for water $\displaystyle =1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ )
(ii)
Write two differences between ideal and non-ideal solutions.
(i)
A solution is prepared by dissolving $\displaystyle 0.025$ g of potassium sulphate in $\displaystyle 2$ L of water at $\displaystyle 27$°C. Assuming potassium sulphate is completely dissociated, determine its osmotic pressure. (Given : $\displaystyle \mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$, Molar mass of $\displaystyle \mathrm{K}_{2} \mathrm{SO}_{4}=174 \mathrm{~g} \mathrm{~mol}^{-1}$ )
(ii)
What type of azeotrope will be formed by a solution of acetone and chloroform ? Give reason. []
Marking-scheme solution
(i)
Vant Hoff factor (i) = $\displaystyle 3$
∆ Tf = i Kf × m
𝛥𝑇𝑓= i K𝑓 .
𝑊𝐵
𝑀𝐵 ×
𝑊𝐴
= $\displaystyle 3$×$\displaystyle 1.86$ ×
= $\displaystyle 1.27$ K
Freezing point of solution
= Freezing point of solvent −∆Tf
= $\displaystyle 273.15$ − $\displaystyle 1.27$ / = $\displaystyle 273$ – $\displaystyle 1.27$
= $\displaystyle 271.88$ K or -$\displaystyle 1.27$ oC / $\displaystyle 271.73$ K
(ii)
Ideal solution
Non ideal solution
It obeys Raoult’s law at all
range of concentration
It does not obey Raoult’s law
∆Hmix = $\displaystyle 0$
Or
∆Vmix = $\displaystyle 0$
∆Hmix ≠ $\displaystyle 0$
Or
∆Vmix ≠ $\displaystyle 0$
(i)
K2SO4 2K+ + SO42−
i = $\displaystyle 3$
πV = i nB R T
π×$\displaystyle 2$ = $\displaystyle 3$ $\displaystyle 0.025$
×
×
π = $\displaystyle 3$ $\displaystyle 0.025$ $\displaystyle 0.082$ $\displaystyle 300$
×
×
×
×
= $\displaystyle 5.3$ ×$\displaystyle 10$ −$\displaystyle 3$ atm
(ii)
Maximum boiling azeotrope
Because this mixture shows negative deviation / Acetone - chloroform interactions are
stronger.
SolutionsColligative Properties and Determination of Molar MassApplylong_answermedium
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CBSE Class 12 Chemistry past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.