The formula for calculating the freezing point depression ΔT_"f"
of an ionic solute is
color(blue)(bar(ul(|color(white)(a/a)ΔT_"f" = iK_text(f)bcolor(white)(a/a)|)))" "
where
ΔT_text(f) = T_text(f)^@ - T_text(f) = the depression of the freezing point
icolor(white)(ml) = the van't Hoff i factor
K_text(f)color(white)(ll) = is the molal freezing point depression constant
bcolor(white)(ml) = is the molal concentration of the solution.
Step 1. Calculate the molal concentration of the solution
A 1 % solution contains 1 g of "Ca"("NO"_3)_2 in 100 g of solution (that is, 1 g solute in 99 g water).
(a) Calculate the moles of "Ca"("NO"_3)_2
"Moles of Ca"("NO"_3)_2 = 1 color(red)(cancel(color(black)("g Ca"("NO"_3)_2))) × ("1 mol Ca"("NO"_3)_2)/(164.09 color(red)(cancel(color(black)("g Ca"("NO"_3)_2)))) = "0.0061 mol Ca"("NO"_3)_2
(b) Calculate the molal concentration of the solution
b = "moles of solute"/"kilograms of solvent" = "0.0061 mol"/"0.099 L" = "0.062 mol/kg"
Step 2. Calculate the freezing point depression
For "Ca"("NO"_3)_2, i = 3 because 1 mol of "Ca"("NO"_3)_2 forms 3 mol of ions
∴ ΔT_"f" = "3 ×1.86 °C"·color(red)(cancel(color(black)("kg·mol"^"-1"))) × 0.062color(red)(cancel(color(black)("mol·kg"^"-1"))) = "0.34 °C"
Step 3. Calculate the freezing point
T_text(f) = T_text(f)^° - ΔT_text(f) = "0.0 °C - 0.34 °C" = "-0.3 °C"
Note: The answer can have only one significant figure because that is all you gave for the concentration of the solution.
