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

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100.52°C

100.10°C

100.26°C

102.6°C

answer is C.

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Detailed Solution

T_b - T_o = 0.52 x 0.1 x 5
T_b - 100 = 0.26
T_b= 100.26°C

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