Question

Which relations are not correct for an aqueous dilute solution of $K_{3}P{O}_4$ if its degree of dissociation is $\alpha$ ?

• A. $\frac{\Delta P}{P^{\circ }}=\frac{\text{ Molality }\times 18\times (1+3\alpha )}{1000}$
• B. $\frac{\Delta P}{P^{\circ }}=\frac{{\pi }_{\text{obs }}\times 18\times (1+3\alpha )}{RT\times 1000}$
• C. $\frac{\Delta P}{P^{\circ }}=\frac{\Delta T_{f\text{ obs }}\times 18}{K_f\times 1000}$
• D. $Mw$ of $K_{3}P{O}_4=M{w}_{\text{obs }}\times (1+3\alpha )$

Answer 1

Answer: (A), (C), (D)

$\frac{\Delta P}{P^{\circ }}=\frac{n_2}{n_1}=\frac{n_2\times M{w}_1\times 1000}{W_1\times 1000}=\frac{\text{ Molality }\times M{w}_1}{1000}$
For electrolyte $\frac{\Delta P}{P^{\circ }}=\frac{\text{ Molality }\times M}{1000}\times (1+3\alpha )$
$(M{w}_1=18$ for $H_{2}0)$
Also, ${\pi }_{obs}=C\times R\times T(1+3\alpha )$
$\therefore \frac{\Delta P}{P^{\circ }}=\frac{{\pi }_{obs}}{RT}\times \frac{18}{1000}$
$\Delta T_f$ obs $=K_f\times$ molaltiy $\times (1+3\alpha )$
$\frac{\Delta P}{P^{\circ }}=\frac{\Delta T_{f\text{ obs }}\times 18}{K_f\times 100}$
$i=(1+3\alpha )=\frac{\text{ Calculated molecular weight }}{\text{ Observed molecular weight }}$
Therefore, molecular weight of $K_{3}P{O}_4=$
$M_{\text{obs }}\times (1+3\alpha )$