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Standard XIII

Chemistry

Depression in Freezing Point

A solution of...

Question

A solution of urea (mol. Mass 60 $g m o l^{- 1}$ ) boils at $100.36^{\circ} C$ at one atmospheric pressure. If the cryoscopic and ebullioscopic constants for water are $1.86$ and $0.512 K k g m o l^{- 1}$ respectively, then the above solution will freeze at :

- 6.54^{\circ} C

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- 0.05^{\circ} C

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- 1.308^{\circ} C

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- 0.654^{\circ} C

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Solution

The correct option is C $- 1.308^{\circ} C$
$△ T_{b} = 0.36 = K_{b} m = 0.512 m \Rightarrow m = \frac{0.36}{0.512}$
so, $△ T_{f} = K_{f} m = 1.86 m = 1.86 \times \frac{0.36}{0.512} = 1.308$
So, the solution will freeze at $- 1.308^{\circ} C$ .

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A solution of urea (mol. Mass 60 g mol−1) boils at 100.36 ∘C at one atmospheric pressure. If the cryoscopic and ebullioscopic constants for water are 1.86 and 0.512 K kg mol−1 respectively, then the above solution will freeze at :

The correct option is C −1.308 ∘C △Tb=0.36=Kbm=0.512 m⇒m=0.360.512 so,△Tf=Kfm=1.86 m=1.86×0.360.512=1.308 So, the solution will freeze at −1.308 ∘C.

A solution of urea (mol. Mass 60 $g m o l^{- 1}$ ) boils at $100.36^{\circ} C$ at one atmospheric pressure. If the cryoscopic and ebullioscopic constants for water are $1.86$ and $0.512 K k g m o l^{- 1}$ respectively, then the above solution will freeze at :

The correct option is C $- 1.308^{\circ} C$
$△ T_{b} = 0.36 = K_{b} m = 0.512 m \Rightarrow m = \frac{0.36}{0.512}$
so, $△ T_{f} = K_{f} m = 1.86 m = 1.86 \times \frac{0.36}{0.512} = 1.308$
So, the solution will freeze at $- 1.308^{\circ} C$ .