Solutions Question 93
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.
Options:
A) 750.6 mmHg
B) 755.8 mmHg
C) 745.98 mmHg
D) 739.56 mmHg
Show Answer
Answer:
Correct Answer: C
Solution:
$ \Delta {p_{theo.}}= $ Lowering in vapour pressure when there is no dissocation $ =p_0\times \frac{wM}{Wm} $ (given, $ p_0=760mm $ , w=14g, W=200gm M=18, m=164) $ =\frac{760\times 14\times 18}{200\times 164}=5.84mm $ Degree of dissociation $ \alpha =\frac{70}{100}=0.7 $
$ Ca{{(NO_3)}_2}\rightarrow C{a^{2+}}+2N{O^{-}}_3 $
$ (n=3) $
$ i=1+(n-1)\alpha $
$ =1+(3-1)0.7=2.4 $ Also, $ i=\frac{\Delta {p_{obs.}}}{\Delta {p_{theo.}}} $
$ =\frac{No\text{. of particles after dissociation}}{No\text{. of particles when there is no dissociation}} $ So. $ \Delta p_{obs}=2.4\times \Delta {p_{theo.}}=2.4\times 5.84 $
$ =14.02mm $
$ p_0-p_{s}=\Delta p_{obs}=14.02 $
$ p_{s}=p_0-14.02=760-14.02=745.98mmHg $
BETA
