Q.

To observe an elevation of boiling point is 0.05oC, the amount of a solute (molecular weight = 100) to be added to 100 g of water (Kb - 0.5) is

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2 g

0.05 g

1 g

0.75 g

answer is C.

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Detailed Solution

Elevation of boiling point,∆Tb=w×Kb×1000M×Win gram(Here, w and W- weights of solute and solvent respectively, M = molecular weight of solute and Kb = constant)On substituting values, we get 0.05=w×0.5×1000100×100 or w=0.05×100×1000.5×1000= 1 g

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