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Classification of Colloids - Based on nature of interaction

Calculate the...

Question

Calculate the amount (in milligrams) of $S e O_{3}^{2 -}$ in solution on the basis of following data:
$20$ mL of $\frac{M}{60}$ solution of $K B r O_{3}$ was added to a definite volume of $S e O_{3}^{2 -}$ solution. The bromine evolved was removed by boiling and excess of $K B r O_{3}$ was back titrated with $5$ mL of $\frac{M}{25}$ solution of $N a A s O_{2}$ . The reactions are given below.
$1) S e O {_{3}}^{2 -} + B r O {_{3}}^{-} + H^{-} \to S e O {_{4}}^{2 -} + B r_{2} + H_{2} O$
$2) B r O {_{3}}^{-} + A s O_{2} + H_{2} O \to B r^{-} + A s O {_{4}}^{3 -} + H^{+}$
(Atomic mass of $K = 39, B r = 80, A s = 75, N a = 23, O = 16, S e = 79$ )

84

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95

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42

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None of the above

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Solution

The correct option is B $95$
The total amount of $K B r O_{3} = 20 \times \frac{1}{60} = 0.333$ millimoles.
Amount of $N a A s O_{2} = 5 \times \frac{1}{25} = 0.2$ millimoles
The amount of $K B r O_{3}$ used for the titration with $N a A s O_{2} = \frac{0.2}{6}$ millimoles.
The amount of $K B r O_{3}$ used for the titration with $S e O_{3}^{2 -} = 0.333$ millimoles $- \frac{0.2}{6}$ millimoles $= 0.3$ millimoles.
The number of moles of $S O_{3}^{2 -}$ originally present in the solution $= 0.3 \times \frac{5}{2} = 0.75$ .
The molar mass of $S e O_{3}^{2 -}$ $= 79 + 3 (16) = 79 + 48 = 127$ g/mol.
The amount of $S O_{3}^{2 -}$ originally present in the solution $= 0.75$ millimoles $\times 127$ g/mol $= 95$ mg.

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Similar questions

Q. Calculate the millimoles of

S e O_{3}^{2 -}

in solution on the basis of following data:

70 m l

\frac{M}{60}

solution of

K B r O_{3}

was added to

S e O_{3}^{2 -}

solution. The bromine evolved was removed by boiling and excess of

K B r O_{3}

was back titrated with

12.5 m L

\frac{M}{25}

solution of

N a A s O_{2}

.
The reactions are given below.
I.

S e O_{3}^{2 -} + B r O_{3}^{-} + H^{+} \to S e O_{4}^{2 -} + B r_{2} + H_{2} O

II.

B r O_{3}^{-} + A s O_{2}^{-} + H_{2} O \to B r^{-} + A s O_{4}^{3 -} + H^{+}

Q. Calculate the mass of

S e O_{3}^{2 -}

in solution on the basis of following data.

20

mL of

\frac{M}{60}

solution of

K B r O_{3}

was added to a definite volume of

S e O_{3}^{2 -}

solution. The bromine evolved was removed by boiling and excess of

K B r O_{3}

was back titrated with

5.1

mL of

\frac{M}{25}

solution of

N a A s O_{2}

. The reactions are given below:

(a)

S e O_{3}^{2 -} + B r O_{3}^{-} + H^{+} ⟶ S e O_{4}^{2 -} + B r_{2} + H_{2} O

(b)

B r O_{3}^{-} + A s O_{2}^{-} + H_{2} O ⟶ B r^{-} + A s O_{4}^{3 -} + H^{+}

70 m l, M / 60 s o l^{n}

K B r O_{3}

in added to

S e O_{3}^{2 -}

. Bromine evolved in removed by boiling & excess of

K B r O_{3}

in back titrated with

12.5 m l M / 25 s o l^{n}

N a A s O_{2}

. Calculate millilitre of

S e O_{3}^{2 -}

Q. Assertion :

B r O_{3}^{-}

shows two reactions :

S e O_{3}^{2 -} + B r O_{3}^{-} + H^{+} ⟶ S e O_{4}^{2 -} + B r_{2} + H_{2} O

A s O_{2}^{2 -} + B r O_{3}^{-} + H^{+} ⟶ B r^{-} + A s O_{4}^{-} + H_{2} O

Reason: The ratio of equivalent masses of

B r O_{3}^{-}

in two reactions is

\frac{5}{6}

Q. To a

25 m l

H_{2} O_{2}

solution, excess of acidified solution of

K I

was added. The iodine liberated required

20 m l

0.3 N

{N a}_{2} S_{2} O_{3}

solution. The volume strength of

H_{2} O_{2}

solution is:

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