Question

Calculate the freezing point of a solution of 500.0 g of ethylene glycol dissovled in 500g water. Kf = 1.86 degrees C/m and Kb (which my instructor said was just extraneous info that is not used here) is 0.512 degrees C/m.

Answer 1

Answer:
`T_(f)` = -29.96 °C

Explanation: A solution has a lower freezing point compared to the pure solvent, due to the solute lowering the vapor pressure of the solvent.

The "new" freezing point is calculated as:

`\Delta T_(f) = K_(f).m`

`K_(f)` is the molal freezing-point depression constant

m is molality concentration, i.e., moles of solute per kilogram of solvent

Before calculating freezing point, let's find moles of ethylene glycol (
`C_(2)H_(6)O_(2)`) in the solution:

Molar mass
`C_(2)H_(6)O_(2)` = 62.08 g/mol

For 500 g:

`n = (500)/(62.08)`

n = 8.05 moles

The molality concentration for 0.5kg of water:

m =
`(8.05)/(0.5)`

m = 16.11

The freezing point will be:

`\Delta T_(f) = K_(f).m`

`\Delta T_(f) = -1.86*16.11`

`\Delta T_(f) = -29.96`

Freezing point of a solution of Ethylene Glycol and Water is -29.96°C