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Depression in Freezing Point

The freezing ...

Question

The freezing point of a solution containing $50 c m^{3}$ of ethylene glycol in 50 g of water is found to be $- 4^{o} C$ . Assuming ideal behaviour, the density of ethylene glycol $(K_{f}$ for water $= 1.86 K k g m o l^{- 1}$ ) is:

1.133 g / c m^{3}

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4.433 g / c m^{3}

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3.235 g / c m^{3}

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None of these

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Solution

The correct option is B $1.133 g / c m^{3}$
Given, ethylene glycol $= 503$

mass of water $= 50 g, K_{f} = 1.86 K m o l^{- 1} k g$

$∴$ mass of glycol $= 50 \times d g$

(where d is the density of glycol)

$Δ T_{f} = 34^{o} C$

Now, $Δ T_{f} = \frac{1000 \times K_{f} \times w}{m \times W}$
or $34 = \frac{1000 \times 1.86 \times 50 \times d}{62 \times 50}$ (Molar mass of glycol $= 62$ )
$∴ d = 1.133 g / c m^{3}$

Hence, the correct option is $A$

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Similar questions

50 {cm}^{3}

50 g

{g/cm}^{3}

(K_{f} for water = 1.86 {K Kg mol}^{- 1})

c m^{3}

- 34^{\circ}

K_{f}

m o l^{- 1}

50 {cm}^{3}

50 g

- 34^{\circ} C

K_{f}

= 1.86 {K kg mol}^{- 1}

50 c m^{3}

50 g

{- 34}^{\circ} C

(K_{f} f o r H_{2} O = 1.86 K k g m o l^{- 1}

Molar mass of Ethylene glycol = 62 g m o l^{- 1}

- 34^{\circ} C

(K_{f} f o r w a t e r = 1.86 K k g m o l^{-} 1)

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