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

##### 1 Answer

Well, I got

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 thefreezing pointof the solventin the context of the solution.#"*"# indicatespuresolvent instead.#i# is thevan'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 thefreezing point depression constantof water.#m# is themolalityof the solution, or the#"mols solute/kg solvent"# .

Ethanol dissolves well in water, but *does not dissociate much at all*, since their

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")#