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Elevation in Boiling Point

An aqueous so...

Question

An aqueous solution of urea freezes at $- {0.186}^{o}$ C. $K_{f}$ for water = $1.86 K k g m o l^{- 1}$ , $K_{b}$ for water = $0.512 K k g m o l^{- 1}$ . The boiling point of urea solution will be :

373.065

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373.186

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373.512

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373.0512

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Solution

The correct option is B $373.0512$ K
Given that,

$K_{f} = 1.86 K k g / m o l; K_{b} = 0.512 K k g / m o l; T_{f, s o l u t i o n} = - {0.186}^{o} C$ and $T_{f, w a t e r} = 0^{o} C$

$Δ T_{f} = 0.186 K$

Since,

$\frac{Δ T_{f}}{K_{f}} = \frac{Δ T_{b}}{K_{b}} = m$

$Δ T_{b} = \frac{0.186}{1.86} \times 0.512 = 0.0512 K$

As, $T_{b, w a t e r} = 373 K,$ and

$Δ T_{b} = 0.0512 K$
Since there is an elevation in boiling point upon addition of solute

$∴ T_{b, s o l u t i o n} = 373 + 0.0512 = 373.0512 K$

Hence option D is correct.

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Similar questions

K_{f} = 1.86 K k g m o l^{- 1}, K_{b} = 0.512 K k g m o l^{- 1}

- {0.186}^{o} C

(K_{f} = 1.86 K k g m o l^{- 1}, K_{b} = 0.512 K k g m o l^{- 1})

{100.15}^{o} C .

K_{f}

K_{b}

m o l^{- 1}

m o l^{- 1}

K_{f} = 1.86

^{- 1}

K_{b} = 0.512

^{- 1}

298 K = 23.756

K_{f} = 1.86 K k g {m o l}^{- 1}]

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