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Standard XII

Chemistry

Ozone

All the three...

Question

All the three oxygen atoms of ozone are utilized in the oxidation of:

K_{2} M n O_{4}

solution

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Moist

I_{2}

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acidified

{F e S O}_{4}

solution

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acidified

{S n C l}_{2}

solution

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Solution

The correct option is D acidified ${S n C l}_{2}$ solution
$3 S n C l_{2} + 6 H C l + O_{3} \to 3 S n C l_{4} + 3 H_{2} O$

In the oxidation of $S n C l_{2}$ to $S n C l_{4},$ all the three oxygen are utilized and no oxygen gas is liberated.

Hence, option D is correct.

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Similar questions

Q. Ozone on reaction with one of the following reagents uses all its oxygen atoms for oxidation. The reagent is:

Q. In Ozone, number of Oxygen atoms are:

Q. Ozone is estimated in air by passing certain volume of air through an acidified or neutral or basic

K I

solution when

O_{2}

is evolved and

I^{⊝}

is oxidised to

I_{2}

. Free

I_{2}

evolved is titrated with standard

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

solution. In an experiment,

10

L of air at

1

atm and

27^{o} C

was passed through an alkaline

K I

solution and

I_{2}

liberated required

2

mL of

\frac{M}{800}

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

solution. The volume percentage of

O_{3}

in sample is :

Q. For estimating ozone in the air, a certain volume of air is passed through an acidified potassium iodide solution. Oxygen is evolved and iodide is then oxidized to give iodine. When such a solution is acidified, free iodine is evolved which can be titrated with standard sodium thio-sulphate solution. In an experiment

10 l i t r e

of air at

1

atm and

27^{o} C

were passed through an alkaline

K I

solution, at the end, the iodine entrapped in the solution on titration as above required

1.5 l i t r e

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

solution. Calculate volume percent of

0. 3

in sample.

Q. The bond orders in the pairs of bonded oxygen atoms in ozone molecules are:

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All the three oxygen atoms of ozone are utilized in the oxidation of:

The correct option is D acidified SnCl2 solution3SnCl2+6HCl+O3→3SnCl4+3H2OIn the oxidation of SnCl2 to SnCl4, all the three oxygen are utilized and no oxygen gas is liberated.Hence, option D is correct.

The correct option is D acidified ${S n C l}_{2}$ solution
$3 S n C l_{2} + 6 H C l + O_{3} \to 3 S n C l_{4} + 3 H_{2} O$

In the oxidation of $S n C l_{2}$ to $S n C l_{4},$ all the three oxygen are utilized and no oxygen gas is liberated.

Hence, option D is correct.