Question

Van’t Hoff’s factor of $\mathrm{aq}.{\mathrm{K}}_2\mathrm{S}{\mathrm{O}}_4$ at infinite dilution has a value equal to

• A. $1$
• B. $2$
• C. $3$
• D. None of these

Answer 1

The correct option is C

$3$

The explanation for the correct option

(C) $\mathbf{3}$

Van't Hoff factor

• The Van't Hoff factor can be used as a measure of how a solute affects related properties, including osmotic pressure, boiling-point elevation, relative vapour pressure lowering, and freezing-point depression.
• It is denoted by "i."
• It is the ratio between the actual concentration of particles produced upon dilution and the concentration of substance before dilution.

$\mathrm{i}=\frac{\mathrm{Number}\mathrm{of}\mathrm{moles}\mathrm{after}\mathrm{dissociation}}{\mathrm{Number}\mathrm{of}\mathrm{moles}\mathrm{before}\mathrm{dissociation}}$

Van’t Hoff’s factor of $\mathbf{aq}\mathbf{.}{\mathbf{K}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}$ at infinite dilution

• The reaction is as follows:
• ${\mathrm{K}}_2\mathrm{S}{\mathrm{O}}_4\to 2{\mathrm{K}}^{+}+{\mathrm{SO}}_4^{2-}$
• In the reaction $1\mathrm{mol}$ potassium sulfate $\left({\mathrm{K}}_2{\mathrm{SO}}_4\right)$ formed $2\mathrm{mol}$ potassium $\left({\mathrm{K}}^{+}\right)$ and $1\mathrm{mol}$ sulfate $\left({\mathrm{SO}}_4^{2-}\right)$ ions.
• The number of moles after dissociation is $3\mathrm{mol}$.
• The number of moles before dissociation is $2\mathrm{mol}$.
• The Van’t Hoff’s factor of $\mathrm{aq}.{\mathrm{K}}_2\mathrm{S}{\mathrm{O}}_4$ at infinite dilution is as follows:

$\begin{array}{rcl}\mathrm{i}& =& \frac{3}{1}\\ & =& 3\end{array}$

• Hence the Van’t Hoff factor is $3$.

The explanation for incorrect options

• Van’t Hoff factor of $\mathrm{aq}.{\mathrm{K}}_2\mathrm{S}{\mathrm{O}}_4$ at infinite dilution is $3$.
• Hence, all the other options (A), (B) and (D) are incorrect.

Therefore the correct option is (C) $\mathbf{3}$.