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Calculate the freezing point of the solution when 31 g of ethylene glycol (C2H6O2) is dissolved in 500 g of water. - CBSE (Science) Class 12 - Chemistry

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Question

Calculate the freezing point of the solution when 31 g of ethylene glycol (C2H6O2) is dissolved in 500 g of water.

(Kf for water = 1.86 K kg mol–1)

Solution

The freezing point of pure water is 273.15 K. On dissolving ethylene glycol, freezing point, being a colligative property, will be lowered.

`DeltaT_f=K_"f"(w_sxx1000)/(M_sxxW`

We are given that

Kf for water=1.86 K kg mol1

Mass of solute, ws=31 g

Molar mass of solute, M12 × × 16 × 2

                                   62 g mol1

Mass of water, W=500 g

`thereforeDeltaT_f=1.86xx(31xx1000)/(62xx500)`

             = 1.86 K

Hence, the freezing point of the solution, 

T273.15 K1.86 K

    = 271.29 K

  Is there an error in this question or solution?
Solution Calculate the freezing point of the solution when 31 g of ethylene glycol (C2H6O2) is dissolved in 500 g of water. Concept: Colligative Properties and Determination of Molar Mass - Depression of Freezing Point.
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