The boiling point of 0.1 m K4[Fe(CN)6] is expected to be (Kb for water = 0.52 K kg mol-1)

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100.52oC

100.10oC

100.26oC

102.6oC

answer is C.

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Detailed Solution

i=1+(n−1)α =1+(5−1)×1 =5ΔTb=i×Kb×m =5×0.52×0.1 =0.26KTs=100+0.26 =100.26oc

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