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?
