$25 \mathrm{~mL}$ of an aqueous solution of $\mathrm{KCl}$ was found to require $20 \mathrm{~mL}$ of $1…

$25 \mathrm{~mL}$ of an aqueous solution of $\mathrm{KCl}$ was found to require $20 \mathrm{~mL}$ of $1 \mathrm{~M} \mathrm{~AgNO}_{3}$ solution when titrated using a $\mathrm{K}_{2} \mathrm{CrO}_{4}$ as indicator. Depression in freezing point of $\mathrm{KCl}$ solution with $100 \%$ ionization will be: $\left[\mathrm{k}_{\mathrm{f}}=2.0^{\circ} \mathrm{mol}^{-1} \mathrm{~kg}ight.$ and $molarity = molality$ $]$
  1. $5.0^{\circ}$
  2. $3.2^{\circ}$
  3. $1.6^{\circ}$
  4. $0.8^{\circ}$

Solution

$M(K C l)=\frac{20 \times 1}{25}=0.8$$\mathrm{M}$
$i=[1+(y-1) x]=2$
$\Delta \mathrm{T}_{\mathrm{f}}=$ molality $\times \mathrm{K}_{\mathrm{f}} \times \mathrm{i}$
$=0.8 \times 2 \times 2=3.2^{\circ}$ ^

Asked in: JEE-TOPICTESTS-CHEMISTRY

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