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 the change in freezing point in#""^@ "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. the effective number of solute particles in solution.#K_f = 1.86^@ "C/m"# is the freezing point depression constant of water. The negative sign is in the equation.#m# is the molality of 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?