25 ml of an aqueous solution of KCl was found to require 20 mL of 1 M $A g N O_{3}$ solution when titrated using a $K_{2} C r O_{4}$ as indicator .The depression in freezing point of $K C l$ solution with 100% ionization will be:
[ $K_{f} = {2.0}^{o} {m o l}^{- 1} k g$ and molarity =molality]

{5.0}^{o}

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{3.2}^{o}

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{1.6}^{o}

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{0.8}^{o}

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Solution

The correct option is B ${3.2}^{o}$
$25 \times M_{(K C l)} = 20 \times 1 M_{(A g N O_{3})}$
$∴ M_{(K C l)} = 0.8$
$Δ T_{f} = M^{^{'}} K_{f} i$ $= 0.8 \times 2 \times 2 = 3.2$

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Similar questions

25 m L

of an aqueous solution of

K C l

was found to require

20 m L

1 M A g N O_{3}

solution when titrated using a

K_{2} C r O_{4}

as indicator. Depression in freezing point of

K C l

solution with

100

% ionisation will be:

[K_{F} = {2.0}^{\circ} m o l^{- 1} k g, m o l a r i t y = m o l a l i t y]

25 mL of an aqueous solution of KCl was found to require 20 mL of 1 $M A g N O_{3}$ solution when titrated using a $K_{2} C r O_{4}$ as indicator. The depression in freezing point of KCl solution with 100% ionization will be:
[ $K_{t} = 2^{o} C$ .kg $m o l^{- 1}$ and molarity = molality]