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Standard XII

Chemistry

Abnormal Colligative Properties

25 mL of an a...

Question

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]

${5.0}^{0} C$

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${3.2}^{0} C$

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${1.6}^{0} C$

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${0.8}^{0} C$

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Solution

The correct option is B
${3.2}^{0} C$

$25 \times M_{(K C l)} = 20 \times 1 M_{(A g N O_{3})} ∴ M_{(K C l)} = 0.8 Δ T_{f} = i k_{f} \times m = 0.8 \times 2 \times 2 = {3.2}^{o}$

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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]

Q. 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]

Q. What will be the depression in freezing point of a

0.5 m

K C l

solution? The molal freezing point constant of water is

{1.86}^{o} C m^{- 1}

0.01 N

solution of

K C l

and

B a C l_{2}

are prepared in water. The freezing point of

K C l

is found to be

- 2^{\circ} C

. What is the freezing point of

B a C l_{2}

solution assuming both

K C l

and

B a C l_{2}

to be completely ionized?

Q. The freezing point of aqueous solution that contains

3

% urea,

7.45

K C l

and

9

% of glucose is: (given

K_{1}

of water

= 1.86

and assume

m o l a r i t y = m o l a l i t y

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25 mL of an aqueous solution of KCl was found to require 20 mL of 1 MAgNO3 solution when titrated using a K2CrO4 as indicator. The depression in freezing point of KCl solution with 100% ionization will be: [Kt=2oC.kg mol–1 and molarity = molality]

The correct option is B 3.20C 25×M(KCl)=20×1M(AgNO3)∴M(KCl)=0.8ΔTf=ikf×m=0.8×2×2=3.2o

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]

The correct option is B
${3.2}^{0} C$

$25 \times M_{(K C l)} = 20 \times 1 M_{(A g N O_{3})} ∴ M_{(K C l)} = 0.8 Δ T_{f} = i k_{f} \times m = 0.8 \times 2 \times 2 = {3.2}^{o}$