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Van't Hoff Factor

An aqueous so...

Question

An aqueous solution freezes at $- {0.36}^{o} C . K_{f}$ and $K_{b}$ for water are $1.8$ and $0.52 K k g m o l^{- 1}$ respectively. Then the value of the boiling point of solution at $1 a t m$ pressure is?

{101.04}^{o} C

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{100.104}^{o} C

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{0.104}^{o} C

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100^{o} C

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Solution

The correct option is B ${100.104}^{o} C$
Given, depression in freezing point $= 0.36$

$K_{b} = 0.52 K k g m o l^{- 1}; K_{f} = 1.8 K k g m o l^{- 1}$

$△ T_{f} = K_{f} m$ where $m$ is the molality

$0.36 = 1.8 \times m$

$m = 0.2$

Elevation in boiling point

$Δ T_{b} = K_{b} \times m$

$Δ T_{b} = 0.52 \times 0.2 = 0.104$

$∴$ New Boiling point $= 100 + 0.104 = {100.104}^{o} C$
Hence, the correct option is $B$

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Similar questions

- {0.93}^{0} C

k_{f} = 1.86 K k g m o l^{- 1}, k_{b} = 0.52 K k g m o l^{- 1}

- {0.186}^{o}

K_{f}

1.86 K k g m o l^{- 1}

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0.512 K k g m o l^{- 1}

K_{f} = 1.86

^{- 1}

K_{b} = 0.512

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298 K = 23.756

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273.0 K

273.0 K

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

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[K_{b} = 0.52 K k g m o l - 1]

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