A 1.2 M aqueous solution of an ionic compound with formula MX has the boiling point of \[101.4^\circ C\]. Calculate the Van't Hoff factor for MX, at the concentration (K= \[0.512^\circ Ckgmo{l^{ - 1}}\]).

Hint: In order to find the Van't Hoff factor, we must first know what a Van't Hoff factor is. Van't Hoff factor (i) is the relationship between the apparent number of moles of solute added in solution determined by the colligative property and the actual number of moles of solute added in order to form a solution.

Complete Solution :
Now let's move onto the given problem. First let us see what a Van't Hoff factor is. Van't Hoff factor (i) is the relationship between the apparent number of moles of solute added in solution determined by the colligative property and the actual number of moles of solute added in order to form a solution. The Van't Hoff factor is the measurement of the deviation from the ideal behavior. In order to find the Van't Hoff factor, we have to make use of the following formula.
\[\Delta {T_b} = i \times {K_b} \times m\]………. (1)
where \[\Delta {T_b}\] is the boiling point elevation. This is one of the colligative properties. When a solute is added the boiling point of the solution will increase. This is known as the boiling point elevation.
- The boiling point of the solution is \[100^\circ C\].
\[\Delta {T_b} = 101.4 - 100 = 1.4^\circ C\]
\[1.4^\circ C\] is the boiling point elevation.
\[{K_b}\] is the ebullioscopic constant and it is given as \[0.512^\circ Ckg mo{l^{ - 1}}\].
m is the molality of the solution and it is defined as the number of moles of solute dissolved in 1kg or 1000g of solvent.
\[m = 1.2\]
Substituting the values in equation (1),
\[\Delta {T_b} = i \times {K_b} \times m\]
\[1.4 = i \times 0.512 \times 1.2\]
\[i = \dfrac{{1.4}}{{0.512 \times 1.2}} = 2.28\]
The Van't Hoff factor i is 2.28.

Note: There are other colligative properties present other than boiling point elevation, they are:
- Relative Lowering of Vapour pressure.
- Depression in freezing point.
- Osmotic pressure.