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Standard XII

Chemistry

Ionization Isomerism

Addition of e...

Question

Addition of excess of $A g N O_{3}$ to an aqueous solution of 1 mole of $P d C l_{2} . 4 N H_{2}$ gives 2 moles of AgCl. The conductivity of this solution corresponds to

1 : 1 electrolyte

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1 : 3 electrolyte

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1 : 2 electrolyte

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1 : 4 electrolyte

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Solution

The correct option is C 1 : 2 electrolyte
1 mole of the $P d$ complex reacts with 2 moles of $A g N O_{3}$ to give 2 moles of $A g C l$ . This means that 2 chloride ions are outside the coordination sphere.
So 3 ions of the complex are formed per molecule of the complex when it is dissolved in water, which means that it is a 1:2 electrolyte.

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Similar questions

When 0.1 mol $C o C l_{3} (N H_{3})_{5}$ is treated with excess of $A g N O_{3},$ 0.2 mol of AgCl are obtained. The conductivity of solution will correspond to
(a) 1:3 electrolyte
(b) 1:2 electrolyte
(c) 1:1 electrolyte
(d) 3:1 electrolyte

Q. In aqueous solution 1 mole of

C o C l_{3} .4 N H_{3}

gives 1 mole of AgCl precipitate with excess

A g N O_{3}

, then what is the secondary valency of metal?

Q. Resistance of a conductivity cell filled with a solution of an electrolyte of concentration 0.1M is 100

Ω

. The conductivity of this solution is 1.29

S m^{- 1}

. Resistance of the same cell when filled with 0.02M of the same solution is 520

Ω

. The molar conductivity of 0.02M solution of the electrolyte will be.
(1)

124 \times 10^{- 4} {S m}^{2}^{- 1}

(2)

1240 \times 10^{- 4} {S m}^{2}^{- 1}

(3)

1.24 \times 10^{- 4} {S m}^{2}^{- 1}

(4)

12.4 \times 10^{- 4} {S m}^{2}^{- 1}

C o C l_{3} .3 N H_{3}

does not form any precipitate with excess

{A g N O}_{3}

solution, whereas 1 mole of

C o C l_{3} .5 N H_{3}

gives two moles of

A g C l

with excess

{A g N O}_{3}

. The van't Hoff factor for both the compounds respectively are:

Q. The molar conductivity of a 1.5M solution of an electrolyte is found to be

138.9 s c m^{2} m o l^{- 1}

. Calculate the conductivity of this solution.

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Addition of excess of AgNO3 to an aqueous solution of 1 mole of PdCl2.4NH2 gives 2 moles of AgCl. The conductivity of this solution corresponds to

Addition of excess of $A g N O_{3}$ to an aqueous solution of 1 mole of $P d C l_{2} . 4 N H_{2}$ gives 2 moles of AgCl. The conductivity of this solution corresponds to