When 0.01 moles of solid HgI2 is completely dissolved in a solution containing 0.02moles of KI in 100g of H2O, the depression in the freezing point of the solution is (Kf for H2O is 1.86 K.Kg mol-1)

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-0.558K

0.558K

-0.744K

0.744K

answer is B.

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Detailed Solution

2KI(aq)+HgI2(s)→K2HgI4(aq)→2K++HgI42−0.02 0.01 0After reaction 0 0 0.01 0.02 0.01Molality of particles of solute = 0.03100×1000ΔTf=1.86×0.3= 0.558 K

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