The amount of ice that will separate out from a solution containing 25 g of ethylene glycol in 100 g of water and is cooled to − 1 0 ∘ C , will be - [Given : Kf for H 2 O = 1.86 K mol − 1 kg ]

Question

The amount of ice that will separate out from a solution containing 25 g of ethylene glycol in 100 g of water and is cooled to − 1 0 ∘ C , will be - [Given : Kf for H 2 ​ O = 1.86 K mol − 1 kg ]

TestHub · Accepted Answer

Given:

Mass of ethylene glycol = 25 g

Molar mass of ethylene glycol (C₂H₆O₂) = $2 \times 12 + 6 \times 1 + 2 \times 16 = 62 g / m o l$

Mass of water = 100 g = 0.1 kg

$K_{f} = 1.86 K k g / m o l$

Observed $Δ T_{f} = 0 - (- 1 0^{\circ} C) = 10 K$

Number of moles of ethylene glycol = $\frac{25 g}{62 g / m o l} = 0.4032 m o l$

Let $W^{'}$ be the mass of water that remains unfrozen (in kg).

Molality $m = \frac{m o l eso f so l u t e}{ma sso f so l v e n t ( k g )} = \frac{0.4032}{W ^{'}}$

Using the depression in freezing point formula:

$Δ T_{f} = K_{f} \cdot m$

$10 = 1.86 \times \frac{0.4032}{W ^{'}}$

$W^{'} = \frac{1.86 \times 0.4032}{10} = 0.0749952 k g \approx 75 g$

Mass of ice separated out = Initial mass of water - Mass of unfrozen water

Mass of ice separated out = $100 g - 75 g = 25 g$

The final answer is $25.0 g$ .

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