Question

Which relations are not correct for an aqueous dilute solution of $K _3 PO _4$ if its degree of dissociation is $\alpha$ ?

• A. $\frac{\Delta P}{P^{\circ}}=\frac{\text { Molality } \times 18 \times(1+3 \alpha)}{1000}$
• B. $\frac{\Delta P}{P^{\circ}}=\frac{\pi_{\text {obs }} \times 18 \times(1+3 \alpha)}{R T \times 1000}$
• C. $\frac{\Delta P}{P^{\circ}}=\frac{\Delta T_{ f \text { obs }} \times 18}{K_{ f } \times 1000}$
• D. $M w$ of $K _3 PO _4=M w_{\text {obs }} \times(1+3 \alpha)$

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 $\frac{\Delta P}{P^{\circ}}=\frac{\text { Molality } \times M}{1000} \times(1+3 \alpha)$
$\left(M w_1=18\right.$ for $\left.H _2 0\right)$
Also, $\pi_{ obs }=C \times R \times T(1+3 \alpha)$
$\therefore \frac{\Delta P}{P^{\circ}}=\frac{\pi_{ obs }}{R T} \times \frac{18}{1000}$
$\Delta T_{ f }$ 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 $K _3 PO _4=$
$M_{\text {obs }} \times(1+3 \alpha)$