Question

$25 \,mL$ of an aqueous solution of $KCl$ was found to require $20\, mL$ of $1\, M$ $AgNO _3$ solution when titrated using $K _2 CrO _4$ as an indicator. What is the depression in freezing point of $KCl$ solution of the given concentration? _________ (Nearest integer).
(Given: $K _f=2.0 \,K \,kg \,mol ^{-1}$ )
Assume
1) $100 \%$ ionization and
2) density of the aqueous solution as $1 \,g \,mL ^{-1}$

Answer 1

At equivalence point,
mmole of $KCl =$ mmole of $AgNO _3$
$=20 \,m$ mole
Volume of solution $=25 \,ml$
Mass of solution $=25\, gm$
Mass of solvent
$=25-$ mass of solute
$=25-\left[20 \times 10^{-3} \times 74.5\right]$
$=23.51 \,gm$
Molality of $KCl =\frac{\text { mole of } KCl }{\text { mass of solvent in } kg }$
$ =\frac{20 \times 10^{-3}}{23.51 \times 10^{-3}}=0.85 $
$ \text { i of } KCl =2(100 \% \text { ionisation }) $
$ \Delta T _{ f }= i \times K _{ f } \times m$
$ =2 \times 2 \times 0.85 $
$=3.4$
$ \approx 3$