# The freezing point of an aqueous solution is -2.79°C. What is the boiling point of this solution?

##### 1 Answer
Jun 16, 2016

 100.765 °C

#### Explanation:

First, you have to find the molality of particles in the solution using the following formula:

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

where:

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

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

$i \to \text{ is the Van'to Hoff factor. It is related to the number of}$
$\text{particles ( or ions) of the solute dissolved in the solution.}$

${K}_{f} \to \text{is the freezing point depression constant of the solvent.}$
"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'= -((-2.79 °C-0°C))/(1.86°C.Kg.mol^-1)

m'= (2.79 cancel (°C))/(1.86cancel (°C.)Kg.mol^-1)

$m ' = 1.5 \setminus \frac{m o l}{K g}$

Once the molality of the solute is determined, the boiling point of the solution could be determined.

The change in the boiling point is determined using the following formula:

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

Where ${K}_{b}$ is the boiling point elevation constant for water.

K_b = 0.510 \°C* Kg*mol^-1

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

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

Delta T= 0.510 \°C* Kg*mol^-1 xx1.5 \ (mol)/(Kg)

Delta T= 0.765 °C

$\Delta T = {T}_{f} - {T}_{i}$

Note that ${T}_{i}$ is the normal boiling point of water and ${T}_{f}$ is the temperature at which the solution boils.

0.765 °C= T_f- 100.000 °C

 T_f=100.765 °C