Question

Van’t Hoff factor of $ H{g_2}C{l_2} $ in aqueous solution will be (it is $ 80\% $ ionised in the solution)
(A) $ 1.6 $
(B) $ 2.6 $
(C) $ 3.6 $
(D) $ 4.6 $

Answer 1

Hint: Van’t Hoff factor is defined as the number of ions that were formed when the given compound or molecule was dissociated in presence of water. The degree of ionisation was represented by $ \alpha $ related to the Van’t Hoff factor as follows and can be determined from that equation.
 $ \alpha = \dfrac{{i - 1}}{{n - 1}} $
 $ \alpha $ is degree of ionisation
 $ i $ is Van’t Hoff factor
 $ n $ is the number of ions formed when dissociated.

Complete answer:
Mercury is the only non-metal that can exist as liquid at room temperature in the periodic table. chlorine is a non-metal with high electronegativity value. Thus, these two atoms can be involved in the formation of a molecule known as mercury salt with the molecular formula of $ H{g_2}C{l_2} $ and the name is mercury chloride.
Given that mercury salt was dissociated in aqueous solution is of $ 80\% $ which means the value of $ \alpha $ is $ 0.8 $ as it is given in percentages and must be converted into a whole number.
The dissociation equation of mercury chloride is $ H{g_2}C{l_2} \rightleftharpoons H{g_2}^{2 + } + 2C{l^ - } $
The total number of ions formed were $ 3 $
Substitute the values in the above formula,
 $ 0.8 = \dfrac{{i - 1}}{{3 - 1}} $
By simplification of the above equation,
 $ i = 0.8\left( 2 \right) + 1 \Rightarrow i = 2.6 $
Thus, the Van't Hoff factor of $ H{g_2}C{l_2} $ in aqueous solution will be $ 2.6 $ .

Note:
Generally, the Van't Hoff factor is the representation of the number of ions that were dissociated in the presence of an aqueous solution. For some salts like mercury chloride, the ionisation of salt is not $ 100\% $ , for these types of salts the number of ions dissociated were not equal to the Van’t Hoff factor.