# What is the freezing point for an aqueous solution with #"0.195 molal"# #"K"_2"S"# dissolved in it? Assume #100%# dissociation. #K_f = 1.86^@ "C/m"#.

##### 1 Answer

Aug 9, 2017

As you should have seen in your book,

#DeltaT_f -= T_f - T_f^"*" = -iK_fm# ,where:

#DeltaT_f# is thechange in freezing pointin#""^@ "C"# , from that of the pure solvent,#T_f^"*"# , to that of the solution,#T_f# .#i# is the van't Hoff factor, i.e. theeffective number of solute particlesin solution.#K_f = 1.86^@ "C/m"# is thefreezing point depression constantof water. The negative sign is in the equation.#m# is themolalityof the solution...#"mol solute/kg solvent"# . What is the solute?

Assuming

#"K"_2"S"(aq) -> 2"K"^(+)(aq) + "S"^(2-)(aq)#

and

#color(blue)(T_f) = T_f^"*" - iK_fm#

#= 0^@ "C" - 3 cdot 1.86^@ "C/m" cdot "0.195 m"#

#= color(blue)ul(-1.088^@ "C")#

What was the change in freezing point?