# Question 03020

Mar 26, 2017

The freezing point is -1.19 °C.

The formula for calculating freezing point depression ΔT_"f" is

color(blue)(bar(ul(|color(white)(a/a)ΔT_"f" = iK_fbcolor(white)(a/a)|)))" "

where

$i$ is the van't Hoff factor

${K}_{\text{f}}$ is the molal freezing point depression constant

$\text{b}$ is the molality of the solution.

In this problem,

$i = 2$, because 1 mol of $\text{NaCl}$ gives 2 mol of particles
${K}_{\textrm{f}} = \text{1.86 °C·kg·mol"^"-1}$
$b = \text{0.321 mol·kg"^"-1}$

ΔT_"f" = 2 × "1.86 °C"·color(red)(cancel(color(black)("kg·mol"^"-1"))) × 0.321 color(red)(cancel(color(black)("mol·kg"^"-1"))) = "1.194 °C"

T_text(f) = T_text(f)^° - ΔT_text(f) = "0 °C - 1.194 °C" = "-1.19 °C"#

Note: The answer can have only 3 significant figures (two decimal places), because that is all you gave for the molality of the solution.