Question

What is the molality for an aqueous solution whose freezing point dropped by `15^@ "C"` due to addition of `"CaCl"_2(s)`? `K_f = 1.86^@ "C/m"` for water.

Answer 1

I got `"2.69 m"`.

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In general, the decrease in freezing point can be found from the following equation:

`bb(DeltaT_f = T_f - T_f^"*")`

`= bb(iK_fm)`

where:

• `T_f` is the freezing point of the `"CaCl"_2(aq)` solution.
• `"*"` means of pure water. That is, `T_f^"*" = 0^@ "C"`.
• `i` is the van't Hoff factor. A good way to estimate it is to say that the number of ions the solute makes is equal to `i`.
• `K_f = -1.86^@ "C/m"` is the freezing point depression constant of water near `0^@ "C"`. The units are read as degrees celcius per molal.
• `m = "mol solute"/"kg solvent"` is the molality of the solution. The unit is read molal.

To solve for the molality:

`m = (T_f^"*" - T_f)/(iK_f)`

Now, we would estimate that `bb(i ~~ 3)` since `"CaCl"_2(aq)` dissociates as follows:

`"CaCl"_2(s) stackrel("H"_2"O"(l))(->) "Ca"^(2+)(aq) + 2"Cl"^(-)(aq)`

giving `bb3` ions in solution. Finally, we know that `T_f^"*" = 0^@ "C"`. So plugging the numbers in gives:

`color(blue)(m) = (-15^@ cancel"C" - 0^@ cancel"C")/(3*(-1.86^@ cancel"C""/m")`

`=` `color(blue)("2.69 m")` (molals)