Question

Which relations are correct for an aqueous dilute solution of $\mathrm{K}_{3} \mathrm{PO}_{4}$ if its degree of dissociation is $\alpha$?
This question has multiple correct options.
A) $\dfrac{\Delta P}{P^{\circ}}=\dfrac{\text { Molality } \times 18 \times(1+3 \alpha)}{1000}$
B) $\dfrac{\Delta P}{P^{\circ}}=\dfrac{\pi_{\mathrm{cbs}} \times 18 \times(1+3 \alpha)}{R T \times 1000}$
C) $\dfrac{\Delta P}{P^{\circ}}=\dfrac{\Delta T_{\text {fobs }} \times 18}{K_{f} \times 1000}$
D) $M w$ of $\mathrm{K}_{3} \mathrm{PO}_{4}=M w_{o b s} \times(1+3 \alpha)$

Answer 1

Hint: We know that an aqueous dilute solution of any salt is formed in water and the molar mass of water is $18\;{gmol}^{-1}$. It is also known that the dissociation of electrolyte takes place in their aqueous solution. Therefore, we can say that the value of $\text{Mw}_{1}\;is\;18\;{gmol}^{-1}$.

Complete step by step answer:
We can write the dissociation reaction of $\mathrm{K}_{3} \mathrm{PO}_{4}$ as follows.
$\mathrm{K}_{3} \mathrm{PO}_{4} \rightarrow 3 \mathrm{K}^{+}+\mathrm{PO}_{4}^{3-}$
We can say that the value of Van't Hoff factor in terms of degree of dissociation $\mathrm{i }=1+3 \alpha$
We can also express the Van't Hoff factor as follows.
$i=\dfrac{\mathrm{M}_{\mathrm{c}}}{\mathrm{M}_{\mathrm{obs}}}=(1+3 \alpha)$
Therefore, we can conclude from the above equation that the molecular weight of $\mathrm{K}_{3} \mathrm{PO}_{4}$ $=M_{obs } \times(1+3 \alpha)$.
We know that the expression for the relative lowering in the vapor pressure is expressed as follows.
${\dfrac{\Delta P}{P^{\circ}} =\dfrac{\mathrm{n}_{2}}{\mathrm{n}_{1}}}$
$=\dfrac{\mathrm{n}_{2} \times \mathrm{Mw}_{1} \times 1000}{\mathrm{W}_{1} \times 1000}$
$=\dfrac{\text {Molality} \times \mathrm{Mw}_{1}}{1000}$
On simplification we can say express the above expression for an electrolyte as follows.
$\begin{aligned} \dfrac{\Delta P}{P^{\circ}} &=\dfrac{\text {Molality} \times \mathrm{Mw}_{1}}{1000} \end{aligned}\times(1+3 \alpha)$
We know that the expression for the observed osmotic pressure is the one which is shown as follows.
${\pi_{\mathrm{obs}} =C R T(1+3 \alpha)}$
$\therefore \dfrac{\Delta P}{P^{\circ}}=\dfrac{\pi_{\mathrm{cbs}}}{R T} \times \dfrac{18}{1000}$
We know that the expression for the observed depression in freezing point is the one which is shown as follows.
${\Delta T_{\text {fobs }} =K_{t} \times \text { molality } \times(1+3 \alpha)}$
$\therefore \dfrac{\Delta P}{P^{\circ}}=\dfrac{\Delta T_{\text {fobs }}}{K_{\mathrm{f}}} \times \dfrac{18}{100}$

Hence, we can conclude that the correct options are A, C and D.

Note: It is known that the value of Van't Hoff factor is always larger than one then the experimental molecular weight will always be less than the theoretical value calculated from the formula of any of the colligative properties. We can also interpret the Van't Hoff factor as the ratio of the number of particles in solution to the number which are obtained to be the assumed zero ionization.