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.

## Expert Answer

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 $2.76$.

## Numerical Answer

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 $2.28$.