#### Question

Calculate the freezing point of the solution when 31 g of ethylene glycol (C_{2}H_{6}O_{2}) is dissolved in 500 g of water.

(K_{f} 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

K_{f} for water=1.86 K kg mol^{−1}

Mass of solute, w_{s}=31 g

Molar mass of solute, M_{s }= 12 × 2 + 1 × 6 + 16 × 2

= 62 g mol^{−1}

Mass of water, W=500 g

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

**= 1.86 K**

Hence, the freezing point of the solution,

T_{f }= 273.15 K−1.86 K

= 271.29 K

Is there an error in this question or solution?

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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. Concept: Colligative Properties and Determination of Molar Mass - Depression of Freezing Point.

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