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Depression in Freezing Point

Calculate the...

Question

Calculate the amount of $C a {C l}_{2}$ (molar mass $= 111 g {m o l}^{- 1}$ ) which must be added to $500 g$ of water to lower its freezing point by $2 K$ , assuming $C a {C l}_{2}$ is completely dissociated. ( $K_{f}$ for water $= 1.86 K k g {m o l}^{- 1}$ )

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Solution

Given,

$Δ T_{f} = 2 K$
$K_{f} = 1.86 K k g m o l^{- 1}$

Using the relation, $Δ T_{f} = m \times K_{f}$

$\Rightarrow 2 = m \times 1.86$
$\Rightarrow m = 1.0752$

Let $x$ g of $C a C l_{2}$ be aded to the solution, hence

$\Rightarrow \frac{\frac{x}{111}}{0.500} = 1.0752$

$\Rightarrow x = 60 g$

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Similar questions

C a C l_{2}

= 111 g m o l^{- 1})

C a C l_{2}

(K_{f}

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

C a C l_{2}

C a C l_{2}

K_{f}

K C l

(k_{f}

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

20.0 g

C H_{3} C H_{2} C H C l C O O H

500 g

K_{a} = 1.4 \times 10^{- 3}, K_{f} = 1.86 K k g m o l^{- 1}]

= 62.0

m o l^{- 1}

5.50

0^{o}

10^{o}

K_{f}

= 1.86

m o l^{- 1}

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