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Q.On dissolving 2.52.5 g of a non-volatile substance in 100100 g of benzene, the boiling point rose by 0.4 ∘C0.4\,^{\circ}C. The molal elevation constant for benzene is 2.672.67. Calculate the molecular weight of the substance.

Uttar Pradesh UpmspUP Board (UPMSP) Intermediate 2020Subjective· 2mImportance★★★★★
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Elevation of boiling point gives M=Kbw 1000ΔTb W≈166.9 g mol−1M=\dfrac{K_b w\,1000}{\Delta T_b\,W}\approx166.9\ \text{g mol}^{-1}.

Concept: Boiling-point elevation is a colligative property: ΔTb=Kb m\Delta T_b=K_b\,m, where the molality m=w×1000M×Wm=\dfrac{w\times1000}{M\times W} (ww = mass of solute, WW = mass of solvent in g, MM = molar mass). Rearranging for MM:

M=Kb w 1000ΔTb W.M=\dfrac{K_b\,w\,1000}{\Delta T_b\,W}.

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