Question

When $20\mathrm{g}$ of naphthoic acid is dissolved in $50\mathrm{g}$ of benzene ${\mathrm{K}}_{\mathrm{f}}=1.72\mathrm{kg}{\mathrm{mol}}^{-1}$ freezing point depression of $2\mathrm{K}$ is observed. The van't Hoff factor is?

Answer 1

Step $1$: Formula for depression of freezing point

• Given data

$$\begin{gathered} {\mathrm{W}}_{\mathrm{B}}(\mathrm{naphthoic}\mathrm{acid})=20\mathrm{g}{\mathrm{W}}_{\mathrm{A}}\left(\mathrm{Benzene}\right)=50\mathrm{g} \\ ∆{\mathrm{T}}_{\mathrm{f}}(\mathrm{Depression}\mathrm{in}\mathrm{freezing}\mathrm{point})=2\mathrm{k} \\ {\mathrm{K}}_{\mathrm{f}}=1.72\mathrm{kg}{\mathrm{mol}}^{-1} \end{gathered}$$

• The formula for depression at the freezing point is:

$∆{\mathrm{T}}_{\mathrm{f}}={\mathrm{K}}_{\mathrm{f}}\frac{{\mathrm{W}}_{\mathrm{B}}}{{\mathrm{M}}_{\mathrm{B}}}\mathrm{x}\frac{1000}{{\mathrm{W}}_{\mathrm{A}}}\mathrm{x}\mathrm{i}$

Where ${\mathrm{M}}_{\mathrm{B}}$ is the molecular weight of naphthoic acid.

Step $2$: Finding the molecular weight of naphthoic acid

• The molecular weight of naphthoic acid is

$$\begin{gathered} {\mathrm{C}}_{11}{\mathrm{H}}_8{\mathrm{O}}_2=(11\mathrm{x}12)+(1\mathrm{x}8)+(2\mathrm{x}16) \\ =172\mathrm{u} \end{gathered}$$

Step $3$: Finding van't Hoff factor from depression in freezing point

• Van't Hoff factor (i) is nothing but the degree of dissociation of solute in the solution.
• Now on substituting the values in the depression of the freezing point formula we get,

$$\begin{gathered} ∆{\mathrm{T}}_{\mathrm{f}}={\mathrm{K}}_{\mathrm{f}}\frac{{\mathrm{W}}_{\mathrm{B}}}{{\mathrm{M}}_{\mathrm{B}}}\mathrm{x}\frac{1000}{{\mathrm{W}}_{\mathrm{A}}}\mathrm{x}\mathrm{i} \\ 2=1.72\frac{20}{172}\mathrm{x}\frac{1000}{50}\mathrm{x}\mathrm{i} \\ \mathrm{i}=\frac{1}{2} \\ \mathrm{i}=0.5 \end{gathered}$$