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elena55 [62]
3 years ago
15

The coolant in automobiles is often a 50/50 % by volume mixture of ethylene glycol, C2H6O2, and water. At 20°C, the density of e

thylene glycol is 1.1088 g/mL and the density of water is 0.9982 g/mL. What is the expected freezing point of a 50/50(v/v)% ethylene glycol/water solution? Kf = 1.86°C/m for water.
Chemistry
1 answer:
Misha Larkins [42]3 years ago
6 0

Explanation:

Let the volume of the solution be 100 ml.

As the volume of glycol = 50 = volume of water

Hence, the number of moles of glycol = \frac{mass}{molar mass}

                                                  = \frac{density \times volume}{molar mass}

                         = \frac{1.1088 \times 50}{62 g/mol}

                         = 0.894 mol

Hence, number of moles of water = \frac{50 \times 0.998}{18}

                                             = 2.77

As glycol is dissolved in water.

So, the molality = 0.894 \times \frac{1000}{49.92}

                           = 17.9

Therefore, the expected freezing point = -1.86 \times 17.9

                                                                  = -33.31^{o}C

Thus, we can conclude that the expected freezing point is -33.31^{o}C.

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