Question

The degree of dissociation of $Ca(NO_{3})_{2}$ in a dilute solution containing $14 \,g$ of the salt per $200 \,g$ of water at $100^{\circ}C$ is $70\%$. If the vapour pressure of water is $760 \,mmHg$, calculate the vapour pressure of solution

• A. $750.6\,mm\,Hg$
• B. $755.8 \,mm\,Hg$
• C. $745.98 \,mm\, Hg$
• D. $739.56 \,mm\, Hg$

Answer 1

Answer: (C)

$\Delta P_{\text{theo}}$ =Lowering in vapour pressure when there is no dissociation
$=P_{0}\times\frac{wM}{Wm}$
(given, $P_{0}=760\,mm, w=14\,g, w=200\,g, M=18, m=164)$
$=\frac{760\times14\times18}{200\times164}=5.84 \,mm $
Degree of dissociation $\alpha=\frac{70}{100}=0.7$
$\underset{(n = 3)}{Ca(NO_{3})_{2}} \rightleftharpoons Ca^{2+}+2NO^{-}_{3}$
$i=1(n-1) \alpha$
$=1+(3-1)0.7=2.4$
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, $\Delta p_{\text{obs}}=2.4\times\Delta p_{\text{theo}}=2.4\times5.84$
$=14.02\,mm$
$p_{0}- p_{s}=\Delta p_{\text{obs}}=14.02$
$p_{s}= p_{0}-14.02=760-14.02$
$=745.98\,mmHg$