Depression in the freezing point

When 0.01 mole of sugar is dissolved in 100g of a solvent, the depression in freezing point is 0.40^o. When 0.03 mole of glucose is dissolved in 50g of the same solvent, depression in the freezing point will be:
(a) 0.60^o (b) 0.80^o (c) 1.60^o (d) 2.40^o

Answer: (d) ΔT_f = mK_f
0.40 = [(0.01 x 1000)/100] x K_f
K_f = 4

again,
T_f = mK_f
= [(0.03 x 1000)/50] x 4
= 2.4

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