Q.

An aqueous solution of glucose boils at 100.01oC. The molal elevation constant for water is 0.5 K mol-1 kg. The number of molecules of glucose in the solution containing 100 g of water is

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6.023×1023

6.023×1022

12.046×1020

12.046×1023

answer is C.

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Detailed Solution

∴ ΔTb=Kb×CmOr Cm=0.010.5=0.02 mole kg−1 solvent∴ number of moles in 100 g water =0.02×1001000 = 0.002 moles of glucose =0.002×6.023×1023 molecules of glucose =2×6.023×1020 glucose moles = 12.046×1020

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