Question

$25mL$ of an aqueous solution of $KCl$ was found to require $20mL$ of $1M$ $AgN{O}_3$ solution when titrated using $K_{2}Cr{O}_4$ as an indicator. What is the depression in freezing point of $KCl$ solution of the given concentration? _________ (Nearest integer).
(Given: $K_f=2.0Kkgmo{l}^{-1}$ )
Assume
1) $100\%$ ionization and
2) density of the aqueous solution as $1gm{L}^{-1}$

Answer 1

Answer: 3

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