Question

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

Answer 1

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 $-33.31^{o}C$.