Question

$9.45g$ of $C{H}_{2}ClCOOH$ is dissolved in $500mL$ of $H_{2}O$ solution and the depression in freezing point of the solution is ${0.5}^{\circ }C$. Find the percentage dissociation.
$(K_f{)}_{H_{2}O}=1.86Kkg{\text{mole}}^{-1}$.
Take
$\text{Density of}{H}_{2}O=1g/mL$

Answer 1

$C{H}_{2}ClCOOH\rightleftharpoons C{H}_{2}ClCO{O}^{-}+H^{+}$
$Initial100$
$\text{Dissociated}\alpha \alpha \alpha$
$\text{Left}(1-\alpha )\alpha \alpha$

$i=\frac{finalmoles}{initialmoles}=\frac{1-\alpha +\alpha +\alpha }{1}=1+\alpha$
$\Delta T_f=i\times K_f\times m$
$0.5=(1+\alpha )\times 1.86\times m$
Molecular mass of $C{H}_{2}ClCOOH$ is 94.5 g
No. of moles of $C{H}_{2}ClCOOH$ is 0.1 mol
$\text{Molality}\left(m\right)=\frac{0.1}{0.5}=0.2$
$W_{H_{2}O}=500g(\text{Density of}{H}_{2}O=1g/mL)$
$0.5=(1+\alpha )\times 1.86\times 0.2$
$\alpha =0.344$
Percentage of dissociation =34.4%