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Q.(a) 18 g of glucose is dissolved in 1kg of water in a saucepan. At what temperature this solution will boil at 1 atm. pressure? Kb for water is 0.52 K kg mol⁻¹ and boiling point is 373.15 K at 1 atm.

(b) What is meant by Cryoscopic Constant?
Uttarakhand UbseUttarakhand Board Intermediate (Class 12) 2026Subjective· 3mImportance★★★★★
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(a) ΔTb=Kb m=0.52×0.1=0.052 K\Delta T_b = K_b\, m = 0.52\times0.1 = 0.052\ \text{K}, so Tb=373.15+0.052=373.20 KT_b = 373.15 + 0.052 = \boxed{373.20\ \text{K}}. (b) The cryoscopic constant is the molal freezing-point-depression constant KfK_f — the drop in freezing point for a 1-molal solution.

Part (a) — boiling point of the glucose solution.

  • Concept: elevation of boiling point (colligative) ΔTb=Kb m\Delta T_b = K_b\, m, where mm = molality.
  • Moles of glucose: n=18180=0.1 moln = \dfrac{18}{180} = 0.1\ \text{mol} (molar mass of glucose, C6H12O6=180 g mol−1\text{C}_6\text{H}_{12}\text{O}_6 = 180\ \text{g mol}^{-1}).
  • Molality: m=0.1 mol1 kg=0.1 mol kg−1m = \dfrac{0.1\ \text{mol}}{1\ \text{kg}} = 0.1\ \text{mol kg}^{-1}.
  • ΔTb=Kb m=(0.52 K kg mol−1)(0.1 mol kg−1)=0.052 K\Delta T_b = K_b\, m = (0.52\ \text{K kg mol}^{-1})(0.1\ \text{mol kg}^{-1}) = 0.052\ \text{K}.
  • Boiling point of solution: Tb=Tb0+ΔTb=373.15+0.052=373.202 K≈373.20 KT_b = T_b^{0} + \Delta T_b = 373.15 + 0.052 = 373.202\ \text{K} \approx \boxed{373.20\ \text{K}}.

Part (b) — cryoscopic constant. …

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