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Molality

11) One gram ...

Question

11) One gram of $N a_{3} A s O_{4}$ is boiled with excess of solid KI in presence of strong HCI. The iodine evolved is absorbed in KI solution and titrated against 0.2 N hypo solution. Assuming the reaction to be $A s O_{4}^{3 -} + 2 H + 2 I \to A s O_{3}^{3 -} + H_{2} O + I_{2}$
calculate the volume of hypo consumed [Atomic weight of As = 75]
(A) 48.1 mL
(B) 38.4 mL

(C) 24.7 mL

(D) 30.3 mL

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Solution

$V o l u m e o f t h i o \equiv V o l u m e o f I_{2} \equiv V o l u m e o f K I \equiv V o l u m e o f N a_{3} A s O_{4} M o l e c u l a r m a s s o f N a_{3} A s O_{4} = 3 \times 23 + 75 + 4 \times 16 = 208 g n f a c t o r = 2 (a s t h e o x i d a t i o n s t a t e c h a n g e d f r o m + 5 t o + 3) E q u i v a l e n t m a s s o f N a_{3} A s O_{4} = \frac{208}{2} = 104 N o r m a l i t y o f N a_{3} A s O_{4} = \frac{1}{104} A s p e r v o l u m e t r i c p r i n c i p l e V_{1} N_{1} = V_{2} N_{2} V o l u m e o f N a_{3} A s O_{4} = 1 L = 1000 m l N o r m a l i t y o f N a_{3} A s O_{4} = \frac{1}{104} V o l u m e o f t h i o = ? N o r m a l i t y o f t h i o = 0.2 N 1000 \times \frac{1}{104} = V_{2} \times 0.2 V_{2} = \frac{1000 \times \frac{1}{104}}{0.2} = 48.0769 m l = 48.1 m l$

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N a_{3} A s O_{4}

is boiled with excess of solid

K I

in the presence of strong

H C l

. The iodine evolved is absorbed in

K I

solution and titrated against

0.2 N

A s O_{4}^{3 -} + 2 H^{+} + 2 I^{-} ⟶ A s O_{3}^{3 -} + H_{2} O + I_{2}

Calculate the volume of thiosulphate hypo consumed, given that the atomic weight of

A s = 75

N a_{2} A s O_{4}

is boiled with excess of solid

K I

in presence of strong

H C l

. The iodine evolved is absorbed in

K I

solution and litrated against

0, 2 N

hypersolution. Assuming the reaction to be

A S O_{4}^{3 -} + 2 H^{+} + 2 I^{-} \to A s O_{3}^{2 -} + H_{2} O + I_{2}

.
Calculate the volume of thiosulphate hypo consumed. [Atomic weight of

A s = 75

2.480 g

K C I O_{3}

are dissolved in conc.

H C I

and the solution was boiled. Chlorine gas evolved in the reactions was then passed through a solution of

K I

and liberated iodine was titrated with

100 m l

12.3 m l

of same hypo solutions required

24.6 m l

0.5

iodine for complete neutralization. Calculate

%

K C I O_{3}

K C l O_{3}

are dissolved in cone. HCl and the solution was boiled. Chlorine gas evolved in the reaction was then passed through a solution of KI and liberated iodine was titrated with 100 mL of hypo. 12.3 mL of same hypo solution required 24.6 mL of 0.5 N iodine for complete neutralization. Calculate % purity of

K C l O_{3}

2.5

g sample of copper is dissolved in excess of

H_{2} S O_{4}

100

0.02 M C u S O_{4} (a q)

10

0.02 M

C u S O_{4} (a q)

is mixed with excess of

K I

to show the following changes:

C u S O_{4} + 2 K I ⟶ K_{2} S O_{4} + C u I_{2}

2 C u I_{2} ⟶ C u_{2} I_{2} + I_{2}

The liberated iodine is titrated with hypo (

N a_{2} S_{2} O_{3}

) which requires

V

0.1 M

hypo solution for its complete reduction.The amount of

I_{2}

liberated in the reaction of

10

0.02 M

solution with excess

K I

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