Question

What is the freezing point depression of a solution containing 100g of ethanol (`C_2H_5OH`) in 0.750 kg of water?

Answer 1

Well, I got `-5.38^@ "C"`, but I only get 1 significant figure, so... `-5^@ "C"`.

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The change in freezing point due to adding solute into a solvent is given by:

`DeltaT_f = T_f - T_f^"*" = -iK_fm`,

where:

• `T_f` is the freezing point of the solvent in the context of the solution. `"*"` indicates pure solvent instead.
• `i` is the van't Hoff factor, i.e. the effective number of solute particles per undissociated particle placed into the solvent.
• `K_f = 1.86^@ "C"cdot"kg/mol"` is the freezing point depression constant of water.
• `m` is the molality of the solution, or the `"mols solute/kg solvent"`.

Ethanol dissolves well in water, but does not dissociate much at all, since their `"pK"_a` values are nearly identical. Thus, `i ~~ 1`.

The mols of ethanol in solution are:

`100 cancel"g EtOH" xx ("1 mol EtOH")/(46.0684 cancel"g EtOH") = "2.17 mols"`

So, the molality is:

`m = "2.17 mols EtOH"/"0.750 kg water"`

`=` `"2.89 mol/kg"`

The change in freezing point is then:

`DeltaT_f = -1 cdot 1.86^@ "C"cdotcancel"kg/mol" cdot 2.89 cancel"mol/kg"`

`= -5.38^@ "C"`

But since you only provided 1 significant figure, we can only say...

`color(blue)(DeltaT_f ~~ -5^@ "C")`