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)
$\Delta T_f=i \times K_f \times m$
$=2 \times 2 \times 0.85$
$=3.4$
$\simeq 3$
