### Redox Reactions Question 8

#### Question 8 - 01 February - Shift 1

$25 mL$ 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

- $100 \%$ ionization and
- density of the aqueous solution as $1 g mL^{-1}$

## Show Answer

#### Answer: (3)

#### Solution:

#### Formula: Molality

#### Formula: Depression in Freezing Point

$ \begin{aligned} & KCl+AgNO_3 \to AgCl+KNO_3 \\ & \downarrow \\ & V=25 ml \quad V=20 ml \\ & M=1 M \end{aligned} $

At equivalence point,

mmole of $KCl=$ mmole of $AgNO_3$

$=20$ mmole

Volume of solution $=25 ml$

Mass of solution $=25 gm$

Mass of solvent

$=25$ - mass of solute

$=25-[20 \times 10^{-3} \times 74.5]$

$=23.51 gm$

Molality of $KCl=\frac{\text{ mole of } KCl}{\text{ mass of solvent in } kg}$

$=\frac{20 \times 10^{-3}}{23.51 \times 10^{-3}}=0.85$

i of $KCl=2$ (100% ionisation)

$\Delta T_f=i \times K_f \times m$

$=2 \times 2 \times 0.85$

$=3.4$

$\simeq 3$