 # A 2.4 m aqueous solution of an ionic compound with the formula MX2 has a boiling point of 103.4 C. Calculate the Van’t Hoff factor (i) for MX2 at this concentration. The aim of this problem is to familiarize us with the calculation of the concentration of an aqueous solution. The concept required to solve this problem is related to molar concentrations, Van’t Hoff factor, and abnormal molar masses.

According to Van’t Hoff’s law, a rise in temperature will result in an expansion in the rate of an endothermic reaction. In order to understand Van’t Hoff law, we must look into Van’t Hoff factor $(i)$, which is the connection between the obvious number of moles of solute mixed in solution specified by the colligative effect and the exact number of moles of solute mixed in order to construct a solution. The formula to calculate $(i)$ is:

$i = \alpha n + (1 – \alpha)$

Where,

$i$ is the Van ‘t Hoff factor,

$\alpha$ is the extent of dissociation, and

$n$ is the number of ions formed during the reaction.

So let’s get on with the given problem. As we have discussed above, the Van’t Hoff factor is basically the measurement of the variation of a solution from its ideal behavior. To calculate the Van’t Hoff factor, we will take help from the following formula:

$\bigtriangleup T_b = i \times K_b \times m……………. (1)$

Where $\bigtriangleup T_b$ is one of the colligative properties responsible for calculating the rise in boiling point. The boiling point of a solution will increase if more solute is added to the solution. This phenomenon is known as boiling point elevation.

We are given the boiling point of the solution $100^{ \circ} C$. Finding $\bigtriangleup T_b$:

$\bigtriangleup T_b = 103.4 – 100 = 3.4^{ \circ} C$

Here, $3.4^{ \circ}C$ is the boiling point elevation.

Whereas $K_b$ is known as the ebullioscopic constant and its value is given as $0.512^{ \circ}C \space kgmol^{-1}$.

And $m$ is the molarity of the solution, defined as the number of moles of solute mixed in $1000g$ of solvent. So:

$m = 2.4$

Substituting the values in equation $(1)$ gives us:

$\bigtriangleup T_b = i \times K_b \times m$

$3.4 = i \times 0.512 \times 2.4$

$i = \dfrac{3.4}{0.512 \times 2.4} = 2.76$

Thus Van’t Hoff factor $i$ is $2.76$.

The Van’t Hoff factor $i$ for $MX_2$ is $2.76$.

## Example

The boiling point of a $1.2 M$ aqueous solution $MX$ is $101.4^{\circ}C$. Find the Van’t Hoff factor for $MX$.

To calculate the Van’t Hoff factor, we will take help from the following formula:

$\bigtriangleup T_b = i \times K_b \times m$

We are given the boiling point of the solution $100^{ \circ} C$. Finding $\bigtriangleup T_b$:

$\bigtriangleup T_b = 101.4 – 100 = 1.4^{ \circ} C$

Here, $1.4^{ \circ}C$ is the boiling point elevation.

$K_b = 0.512^{ \circ}C \space kgmol^{-1}$.

And $m = 1.2$.

Substituting the values in the equation of $T_b$ gives us:

$1.4^{\circ}C = i \times 0.512^{\circ}C\space kgmol^{-1} \times 1.2$

$i = \dfrac{1.4}{0.512 \times 1.2} = 2.28$

Thus, the Van’t Hoff factor $i$ is $2.28$.