Question

A solution of urea in water has a boiling point of 373.128k. Calculate the freezing point of the same solution.

Answer 1

Answer:

272.542 K

Explanation:

Given boiling point of urea = 373.128 K

ΔTb = 373.128–373 = 0.128 K =kb*m

we know the formula for change in  boiling point = kb*m

here k = cryoscopic constant

We know that : kb =0.52 K/m; kf = 1.86K/m.

m = 0.128/0.52 =0.246 m.

now formula for change in freezing point = kf*m

putting the value we get

ΔTf = kf*m = 1.86*0.246 K = 0.458 K

Thus Tf = (273–0.458) = 272.542 K

Answer 2

Answer:

Freezing point = 272.542 K.

Explanation: Given:

B.P = 373.128

Kf = 1.86K/m

Kb=0.52K/m

We need to find Tf

Tf = 273- ΔTf

ΔTf = Kf*m

m= ΔTb/Kb

ΔTb = 373.128-373 = 0.128 = Kb*m

m= 0.128/0.52

m=0.246

Putting the value of 'm' in our fomula

ΔTf = 1.86*0.246

ΔTf = 0.458

Thus Tf = (273–0.458) = 272.542 K.

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