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Q.What is meant by elevation of boiling point? Boiling point of water is 100°C. Calculate the boiling point of an aqueous solution containing 5 g urea (Mol. Mass = 60) in 100 g water. (Kb for water = 0.52 K.kg.mol^-1) [2+3=5] OR Deduce an expression for the rate constant of a zero order reaction. A first order reaction completes 99% in 32 min. Calculate the rate constant of the reaction. [3+2=5]

Odisha ChseOdisha CHSE +2 Science Board Exam 2026Subjective· 5mImportance★★★★★
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Part 1: boiling point of the urea solution ≈ 100.43°C. OR Part 2: zero order k=([A]0−[A])/tk=([A]_0-[A])/t; the given first-order reaction has k≈0.144 min−1k\approx0.144\ min^{-1}.

Elevation of boiling point + calculation:

When a non-volatile solute is dissolved in a solvent, the vapour pressure of the solution is lowered, and consequently the solution must be heated to a HIGHER temperature than the pure solvent's boiling point before its vapour pressure equals atmospheric pressure. This increase in boiling point, ΔTb=Tb(solution)−Tb(solvent)\Delta T_b = T_b(\text{solution}) - T_b(\text{solvent}), is called elevation of boiling point, and for a dilute solution it is proportional to molality: ΔTb=Kb×m\Delta T_b = K_b \times m.

Calculation: moles of urea =560=0.0833 mol= \dfrac{5}{60} = 0.0833\ mol. Mass of water =100 g=0.1 kg= 100\ g = 0.1\ kg.

m=0.08330.1=0.833 mol/kgm = \frac{0.0833}{0.1} = 0.833\ mol/kg

ΔTb=Kb×m=0.52×0.833=0.433 K\Delta T_b = K_b \times m = 0.52 \times 0.833 = 0.433\ K

Tb(solution)=100+0.433=100.43∘CT_b(\text{solution}) = 100 + 0.433 = 100.43^{\circ}C

OR — Zero order rate constant + first-order calculation: …

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