Question

Which relations are correct for an aqueous dilute solution of $K_{3}P{O}_4$ if its degree of dissociation is $\alpha$?

• A. $\frac{\Delta P}{P^o}=\frac{Molality\times 18\times (1+3\alpha )}{1000}$
• B. $\frac{\Delta P}{P^o}=\frac{{\pi }_{obs}\times 18\times (1+3\alpha )}{RT\times 1000}$
• C. $\frac{\Delta P}{P^o}=\frac{\Delta T_{fobs}\times 18}{K_f\times 1000}$
• D. $Mwof{K}_{3}P{O}_4=M{w}_{obs}\times (1+3\alpha )$

Answer 1

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

$\frac{\Delta P}{P^o}=\frac{\Delta T_{fobs}\times 18}{K_f\times 1000}$
$\frac{\Delta P}{P^o}=\frac{Molality\times 18\times (1+3\alpha )}{1000}$
$Mwof{K}_{3}P{O}_4=M{w}_{obs}\times (1+3\alpha )$

$\frac{\Delta P}{P^o}=\frac{n_2}{n_1}=\frac{n_2\times M{w}_1\times 1000}{W_1\times 1000}=\frac{Molality\times M{w}_1}{1000}$

For electrolyte $\frac{\Delta P}{P^o}=\frac{Molality\times M}{1000}\times (1+3\alpha )$
$(M{w}_1=18for{H}_{2}O)$

Also,${\pi }_{obs}=C\times R\times T(1+3\alpha )$

$\therefore \frac{\Delta P}{P^o}=\frac{{\pi }_{obs}}{RT}\times \frac{18}{1000}$

$\Delta T_{fobs}=K_f\times molality\times (1+3\alpha )$

$\frac{\Delta P}{P^o}=\frac{\Delta T_{fobs}\times 18}{K_f\times 100}$

$i=(1+3\alpha )=\frac{Calculatedmolecularweight}{Observedmolecularweight}$

Therefore, molecular weight of $K_{3}P{O}_4=M_{obs}\times (1+3\alpha )$

Hence, options $A,C$ and $D$ are correct relations