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.