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:

1 : 3 electrolyte

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1 : 2 electrolyte

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1 : 1 electrolyte

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3 : 1 electrolyte

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Solution

The correct option is B 1 : 2 electrolyte
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.
When 1 mol $C o C l_{3} (N H_{3})_{5}$ is treated with excess of $A g N O_{3}$ , 2 mol of AgCl will be obtained.
Hence, one molecule of $C o C l_{3} (N H_{3})_{5}$ dissociates to form two chloride ions. $C o C l_{3} (N H_{3})_{5}$ can be represented as $[C o (N H_{3})_{5} C l] C l_{2}$ .
$[C o (N H_{3})_{5} C l] C l_{2} \to [C o (N H_{3})_{5} C l]^{2 +} + 2 C l^{-}$ .

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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. When

0.1

m o l

C o {C l}_{3} {(N H_{3})}_{5}

is treated with excess of

A g N O_{3},

0.2

mole of

A g C l

are obtained. The conductivity of solution will correspond to

Q. When

0.1 m o l C o C l_{3} (N H_{3})_{5}

is treated with excess of

A g N O_{3}, 0.2 m o l

A g C l

are obtained. The conductivity of solution will correspond to:

Q. When 1 mol

C r C l_{3} .6 H_{2} O

is treated with excess of

A g N O_{3}

, 3 mol of AgCl are obtained. The formula of the complex is:

Q. How many moles of

A g C l

would be obtained, when 100 ml of 0.1 m

C o C l_{3} (N H_{3})_{5}

is treated with excess of

A g N O_{3}

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When 0.1 mol CoCl3(NH3)5 is treated with excess of AgNO3, 0.2 mol of AgCl are obtained. The conductivity of solution will correspond to:

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: