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Elevation in Boiling Point

An aqueous so...

Question

An aqueous solution freezes at $- {0.186}^{o} C$ $(K_{f} = 1.86 K k g m o l^{- 1}, K_{b} = 0.512 K k g m o l^{- 1})$ . The evalation of boiling point of the solution is____________.

0.186

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0.512

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0.275

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0.0512

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Solution

The correct option is D $0.0512$
$△ T = K_{f} \times m$
$0.186 = 1.86 \times m$
$m = 0.1$
$△ T = K_{b} \times m = 0.512 \times 0.1 = 0.0512$

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Similar questions

- {0.186}^{o}

K_{f}

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

K_{b}

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

{0.186}^{\circ} C

K_{f}

m o l e^{- 1}

K_{b}

m o l e^{- 1}

{0.186}^{o} C

K_{f}

= {1.86}^{o} C k g m o l^{- 1}

K_{b}

= {0.512}^{o} C k g m o l^{- 1}

{100.17}^{o} C .

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

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

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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