Question

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"`.

Answer 1

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 `100%` dissociation...

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

and `1 + 2 = 3 ~~ i`, so...

`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?