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

Step-by-step 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 =
`(mass)/(molar mass)`

=
`(density * volume)/(molar mass)`

=
`(1.1088 * 50)/(62 g/mol)`

= 0.894 mol

Hence, number of moles of water =
`(50 * 0.998)/(18)`

= 2.77

As glycol is dissolved in water.

So, the molality =
`0.894 * (1000)/(49.92)`

= 17.9

Therefore, the expected freezing point =
`-1.86 * 17.9`

=
`-33.31^(o)C`

Thus, we can conclude that the expected freezing point is
`-33.31^(o)C`.