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Depression in Freezing Point

45 g of ethy...

Question

$45 g$ of ethylene glycol is mixed with $600 g$ of water. What is the freezing point of the solution? $k_{f} = 1.86 K k g m o l^{- 1}$ .

- 270.90 K

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270.90

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273 K

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274.15 K

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Solution

The correct option is C $270.90$
45 g of ethylene glycol is mixed with 600 g of water.
Molar mass of ethylene glycol $= 62$ g/mol
45 g ethylene glycol $= \frac{45}{62} = 0.726$ moles.
600 g water $= 0.600$ kg
Molality $(m) = \frac{0.726 m o l}{0.600 k g} = 1.21 m$
The depression in the freezing point $Δ T_{f} = K_{f} m = 1.86 \times 1.21 = 2.25 K$
The freezing point of pure water $= 273.15$ K.
Freezing point of solution $= 273.15 + 2.25 = 270.90$ K
Hence, the option (B) is the correct answer.

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