Question

If relative decrease in vapour pressure is $0.4$ for a solution containing $1$ mol $NaCl$ in $3$ mol $H_{2}O$. $NaCl$ is ______ % ionised.

• A. $60$%
• B. $50$%
• C. $100$%
• D. $40$%

Answer 1

Answer: (A)

$60$%

Relative decrease in vapour pressure $\frac{p_{solvent}^0-p_{solvent}}{p_{solvent}^0}=i\times {\chi }_{solute}$

${\chi }_{solute}=\frac{molesofsolute}{molesofsolute+molesofsolvent}$

Given that $n_{NaCl}=1$ and $n_{H_{2}O}=3$ and $\frac{p_{solvent}^0-p_{solvent}}{p_{solvent}^0}=0.4$

Thus $0.4=i\times \frac{1}{1+3}$

$i=4\times 0.4=1.6$

Let the degree of dissociation of NaCl be $\alpha$=fraction of molecules of NaCl dissociated=number of moles of NaCl dissociated/initial moles of NaCl present

For 1mol of NaCl:

$NaCl\to N{a}^{+}+C{l}^{-}$

$1-\alpha \alpha \alpha$

total number of moles after dissociation =$1-\alpha +\alpha +\alpha =1+\alpha$

since $i=effectivenumberofmolecules/initialnumberofmolecules$

$i=(1+\alpha )/1$

as above calculated, i=1.6

thus$1.6=\frac{1+\alpha }{1}$

$\alpha =1.6-1=0.6$

degree of dissociation =0.6

and therefore % dissociation = $0.6\times 100$

= $60$ %