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Avogadro's Law & Gay Lussac's Law

An aqueous so...

Question

An aqueous solution freezes at $- {0.186}^{o} C (K_{f} = {1.86}^{o}; K_{b} = {0.512}^{o})$ . What is the elevation in boiling point?

0.186

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0.512

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\frac{0.512}{1.86}

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0.0512

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Solution

The correct option is D $0.0512$
$Δ T_{b} = K_{b} \times Molality$

$Δ T_{f} = K_{f} \times Molality$

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

$\Rightarrow Δ T_{b} = \frac{Δ T_{f} \times K_{b}}{K_{f}} = \frac{0.186 \times 0.512}{1.86} = 0.0512$

Hence, the correct option is $C$

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Similar questions

{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}

- {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})

- {0.186}^{0} C

K_{f} = {1.86}^{0} m o l^{- 1} k g

K_{b} = {0.512}^{0} m o l^{- 1} k g

N a C l

- {0.816}^{\circ} C

K_{b (H_{2} O)} = 0.512

^{- 1}

K_{f (H_{2} O)} = 1.86

^{- 1}

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