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: $K _f=20 \,K \,kg \,mol ^{-1}$ ) 

Assume 1) $100 \%$ ionization and 

2) density of the aqueous solution as $1 \,g \,mL ^{-1}$

Answer 1

Correct Answer: 3

The correct answer is 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 $KCl=\frac{\text{mole of}KCl}{\text{mass of solvent in}kg}$
$=\frac{20\times 10^{-3}}{23.51\times 10^{-3}}=0.85$
$\text{i of}KCl=2(100\%\text{ionisation})$
$\Delta T_f=i\times K_f\times m$
$=2\times 2\times 0.85$
$=3.4$
$\approx 3$