# The freezing point of 1% aqueous solution of calcium nitrate will be?

Jun 19, 2018

-0.3 °C

#### Explanation:

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
$i \textcolor{w h i t e}{m l} =$ the van't Hoff $i$ factor
${K}_{\textrm{f}} \textcolor{w h i t e}{l l} =$ is the molal freezing point depression constant
$b \textcolor{w h i t e}{m l} =$ 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

$b = \text{moles of solute"/"kilograms of solvent" = "0.0061 mol"/"0.099 L" = "0.062 mol/kg}$

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.