Question

The degree of dissociation of $Ca(N{O}_3{)}_2$ in a dilute solution containing $14g$ of the salt per $200g$ of water at $100^{\circ }C$ is $70\%$. If the vapour pressure of water is $760mmHg$, calculate the vapour pressure of solution

• A. $750.6mmHg$
• B. $755.8mmHg$
• C. $745.98mmHg$
• D. $739.56mmHg$

Answer 1

Answer: (C)

$\Delta P_{\text{theo}}$ =Lowering in vapour pressure when there is no dissociation
$=P_0\times \frac{wM}{Wm}$
(given, $P_0=760mm,w=14g,w=200g,M=18,m=164)$
$=\frac{760\times 14\times 18}{200\times 164}=5.84mm$
Degree of dissociation $\alpha =\frac{70}{100}=0.7$
$\underset{(n=3)}{Ca(N{O}_3{)}_2}\rightleftharpoons C{a}^{2+}+2N{O}_3^{-}$
$i=1(n-1)\alpha$
$=1+(3-1)0.7=2.4$
Also, $i=\frac{\Delta P_{\text{obs}}}{\Delta P_{\text{theo}}}=\frac{\text{No. o f particles after dissociation}}{\text{No.of particles when there is no dissociation}}$
So, $\Delta p_{\text{obs}}=2.4\times \Delta p_{\text{theo}}=2.4\times 5.84$
$=14.02mm$
$p_0-p_s=\Delta p_{\text{obs}}=14.02$
$p_s=p_0-14.02=760-14.02$
$=745.98mmHg$