1.00 g of a non-electrolyte solute (molar mass 250 g mol-1) was dissolved in 51.2 g of benzene. If the freezing point depression constant, Kf of benzene is 5.12 K kg mol-1, the freezing point of benzene will be lowered by
(2) 0.4 K
Tf = Kf w2 1000M2 w1
Tf = 5.12 1.00 1000250 51.2
= 512012800
= 0.4 K
Answer: (2)
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