Question

$25mL$ of an aqueous solution of $KCl$ was found to require $20mL$ of $1MAgN{O}_3$ solution when titrated using a $K_{2}Cr{O}_4$ as indicator. Depression in freezing point of $KCl$ solution with $100$% ionisation will be:
$[K_F={2.0}^{\circ }mo{l}^{-1}kg,molarity=molality]$

• A. ${3.2}^{\circ }$
• B. ${1.6}^{\circ }$
• C. ${0.8}^{\circ }$
• D. ${5.0}^{\circ }$

Answer 1

The correct option is A ${3.2}^{\circ }$

During neutralisation,

$M_1{V}_1=M_2{V}_2$ since n-factor is 1 in both the cases

Now,

$M_1=\frac{1\times 20}{25}=0.8M$

Again,

$i=(1+x)=1+1=2$

$\therefore △T_f=molality\times K_f\times i$

$=0.8\times 2\times 2$

$={3.2}^{\circ }C$

[Note: $Molality\approx Molarity$]

$\therefore T_s=T_f+\Delta T_f=0+3.2={3.2}^{o}C$