Question

When 20 g of naphthoic acid $\left( {{{\rm{C}}_{{\rm{11}}}}{{\rm{H}}_{\rm{8}}}{{\rm{O}}_{\rm{2}}}} \right)$ is dissolved in 50 g benzene $\left( {{K_f} = 1.72\,{\rm{K}}\,{\rm{kg}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}} \right)$ freezing point depression of 2K is observed. The van’t Hoff factor (i) is:
A.0.5
B.1
C.2
D.3

Answer 1

Hint: We know that, van’t Hoff factor was introduced in 1880 by van’t Hoff to account for association or dissociation. It is represented by the symbol ‘i’. The formula of van’t Hoff factor is, $i = \dfrac{{{\rm{Normal}}\,\,{\rm{molar}}\,{\rm{mass}}}}{{{\rm{Abnormal}}\,\,{\rm{molar}}\,{\rm{mass}}}}$. Here, we have to use the expression of freezing point depression comprising van’t Hoff factor, that is, $\Delta {T_f} = i{K_f}m$

Complete step by step answer:
 Addition of the van’t Hoff factor to modify the equation of freezing point depression.

$\Delta {T_f} = i{K_f}m$ …… (1)

Here, $\Delta {T_f}$ is freezing point depression, I is van’t Hoff factor, ${K_f}$ is freezing constant and m is molality.

First we have to calculate the molality of naphthoic acid. The formula of molality is,

Molality=$\dfrac{{{\rm{Moles}}\,{\rm{of}}\,{\rm{solute}}}}{{{\rm{Volume}}\,{\rm{of}}\,{\rm{solvent}}\,{\rm{in}}\,{\rm{Kg}}}}$ …… (2)

To calculate moles of solute the formula is,

${\rm{Moles}}\,{\rm{of}}\,{\rm{solute}} = \dfrac{{{\rm{Mass}}}}{{{\rm{Molar}}\,{\rm{mass}}}}$

Here, mass of naphthoic acid $\left( {{{\rm{C}}_{{\rm{11}}}}{{\rm{H}}_{\rm{8}}}{{\rm{O}}_{\rm{2}}}} \right)$ is 20 g and molar mass of naphthoic acid is \[ = 11 \times 12 + 8 \times 1 + 16 \times 2 = 172\,{\rm{g}}\]

So, moles of naphthoic acid=$\dfrac{{20}}{{172}}$

Now, we have to use equation (2) to calculate molality. Volume of solvent is 50 g
So, molality of the solution =$\dfrac{{\dfrac{{20}}{{172}}}}{{\dfrac{{50}}{{1000}}}} = 2.32$

The freezing point depression $\left( {\Delta {T_f}} \right)$ is 2K and value of ${K_f}$ is $1.72\,{\rm{K}}\,{\rm{kg}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}$ and molality is 2.32. Put all these values in equation (1).

$\Delta {T_f} = i{K_f}m$

$ \Rightarrow 2 = i \times 1.72\, \times 2.32$

$ \Rightarrow i = \dfrac{2}{{1.72 \times 2.32}}$

$ \Rightarrow i = 0.5$

Therefore, the van’t Hoff factor is 0.5

So, the correct answer is Option A.

Note: We know,

$i = \dfrac{{{\rm{Normal}}\,\,{\rm{molar}}\,{\rm{mass}}}}{{{\rm{Abnormal}}\,\,{\rm{molar}}\,{\rm{mass}}}}$

Here, the molar mass obtained through the experiment is termed as abnormal molar mass.

Also we can write above equation as

$i = \dfrac{{{\rm{Observed}}\,{\rm{colligative}}\,{\rm{property}}}}{{{\rm{Calculated colligative}}\,{\rm{property}}}}$


The calculated colligative properties are obtained by assuming that there is neither association nor dissociation of non-volatile solutes.

Or

$i{\rm{ = }}\dfrac{{{\rm{Total}}\,\,{\rm{number}}\,{\rm{of}}\,{\rm{moles}}\,{\rm{of}}\,{\rm{particles}}\,{\rm{after}}\,{\rm{association}}\,{\rm{or}}\,{\rm{dissociation}}}}{{{\rm{Number}}\,{\rm{of}}\,{\rm{moles}}\,{\rm{of}}\,{\rm{particles}}\,{\rm{before}}\,{\rm{association}}\,{\rm{or}}\,{\rm{dissociation}}}}$

Always remember that the value of van’t Hoff factor is less than one in case of association and more than one in case of dissociation. We know that KCl undergoes dissociation so its value