# The freezing point of an aqueous sodium chloride solution is -0.20°C. What is the molality of the solution?

Jun 16, 2016

$0.055 \setminus \frac{m o l}{K g}$

#### Explanation:

$\Delta T = {K}_{f} \times m \times i$

where:

$m \to \text{ is the molality of the solute expressed in } \frac{m o l}{K g}$

$\Delta T \to \text{ is the difference between the freezing point of the }$
$\text{solution"\ (T_f)" and that of the pure solvent } \left({T}_{i}\right) .$
$\setminus \Delta T = {T}_{f} - {T}_{i}$

$i \to \text{ is the Van'to Hoff factor. It is related to the number of}$
$\text{particles ( or ions) of the solute dissolved in the solution.}$
$\text{in the case of " NaCl " " i = 2 " since " NaCl" fully ionizes into 2 ions}$

${K}_{f} \to \text{is the freezing point depression constant of the solvent.}$
"in aqueous solutions the solvent is water and " K_f = 1.86\°C*Kg*mol^-1

$\Delta T = - {K}_{f} \setminus \times {\underbrace{m \times i}}_{\text{molality of particles}}$

$\Delta T = - {K}_{f} \setminus \times {\underbrace{m \times i}}_{m '}$

$\Delta T = - {K}_{f} \times \setminus m '$

$m ' = - \frac{\Delta T}{K} _ f$

$m ' = - \frac{\left({T}_{f} - {T}_{i}\right)}{K} _ f$

m'= -((-0.20 °C-0°C))/(1.86°C.Kg.mol^-1)

m'= (0.20 \ cancel (°C))/(1.86cancel (°C.)Kg.mol^-1)

$m ' = 0.11 \setminus \frac{m o l}{K g}$

Note that $m '$ is the molality of the dissolved particles in the solution i.e $\setminus N {a}^{+} + C {l}^{-} 1$ while $m$ is the molality of the solute ($N a C l$)

$m \times i = m '$

$m = \frac{0.11}{2} \setminus \frac{m o l}{K g}$

$m = 0.055 \frac{m o l}{K g}$