Find the freezing point (in `""^@ "C"`) for a solution consisting of `"60.0 g"` water and `"40.0 g"` ethylene glycol?

The freezing point depression for water is -1.85 degrees per molal solution.

A molal solution is l mole of particles per kg of solution.

As ethylene glycol is a covalent molecule it does not ionize in water. Because of this 1 mole of ethylene glycol produces 1 mole of particles.

To find the number of moles of ethylene glycol divide the mass of ethylene glycol by the molar mass.

`40 / 62` = ,645 moles.

There are 1000 grams of water in a Kg. To find the number of Kg of solution define the 60 grams grams of water by 1000 grams.

`60 g / 1000 g` = .06 Kg.

The molal concentration is the moles divided by the Kg.

,645 moles / .06 Kg = 10.8 m ( molal concentration)

The freezing point depression is -1.85 `C^o` per 1 -molal.
To find the depression multiply -1.85 x 10.8 molal.

`-1.85xx 10.8` = - 19.88

The normal freezing point is `0^o C` so the new freezing point is
0 + - 19.88 = - 19.88

The closest answer is B.

I got `"b"`.

`a)` is incorrect because freezing point depression is negative.
`c)` is incorrect because it results from using `K_b` instead of `K_f`, and yet the change in boiling point is added to the freezing point...
`d)` is incorrect because it results from having calculated the molality for water instead of for ethylene glycol, and then used `K_b` instead of `K_f`, and then still added the change in boiling point to the freezing point.

The `K_f` of water is `"1.86"^@ "C/m"`, and recall the freezing point depression equation:

`bb(DeltaT_f = -iK_fm)`

where:

• `i` is the van't Hoff factor, which is the number of ions in a fully ionic compound. This was not given so we have to work from a different equation or estimate `i`.
• `K_f` is the freezing point depression constant.
• `m` is the molality of the solution, i.e. `"mols solute"/"kg solvent"`.
• `DeltaT_f = T_f - T_f^"*"` is the change in freezing point, where `T_f^"*"` is the freezing point of the pure solvent and `T_f` is the new freezing point.

Ethylene glycol is not fully ionic, but it is polar, so it is miscible in water. However, its `"pKa"` is significantly higher than that of water (`25` vs. `15.7`), so we can say it does not dissociate significantly in water. Therefore, `i ~~ 1`. It's probably a little higher than `1` for a real solution, however.

The molality of the solution is based on which substance is the solvent, i.e. which one there is more of.

`"60.0 g"/"18.015 g/mol" = "3.331 mol water"`
`"40.0 g"/"62.0 g/mol" = "0.6452 mol EG"`

Therefore, water is the solvent.

`color(green)(m_"soln") = "mols EG"/"kg water"`

`= "0.6452 mols EG"/(60.0 cancel"g water" xx "1 kg water"/(1000 cancel"g")`

`=` `color(green)("10.75 m solution")`

Therefore, the change in freezing point is:

`DeltaT_f\ = T_f - T_f^"*" ~~ (1)("1.86"^@ "C"cdotcancel"kg/mol")(10.75 cancel"mol/kg") ~~ -20.0^@ "C"`

Since the freezing point must decrease, the final freezing point is:

`color(blue)(T_f) = - 20.0^@ "C" + T_f^"*"`

`= color(blue)(-20.0^@ "C")`

The apparent answer is therefore `-20.1^@ "C"`, which is `"B"`. None of the other answers are right because they are positive and not negative.

Based on this answer, can you see that `i ~~ 1.005`? It was a good estimate to say that `i ~~ 1`, but it's not exactly `1`.