Question

calculate the f.p.of aquaous solution containing 10.5 gm of glucose in 200gm of water assume the complete dissociation of MgBr_2 ( molar mass=184) kf = 1.86​

Answer 1

Explanation:

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Answer 2

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Given data,

W₂ = 10.5 g

W₁ = 200 g

Molar mass of MgBr, M₂ = 184 g

Kf of water = 1.86

Hence the change in freezing point is given by the relation,

ΔTf = (1000 x Kf x W₂)/(W₁ x M₂)

= (1000 x 1.86 x 10.5)/(200 x 184)

=> ΔTf = 0.53 K

Hence new freezing point ,

Tf = T₀ - ΔTf = 273 - 0.53 = 272.47 K

Hence the freezing point of the solution will be 272.47 K

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