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Standard XII

Chemistry

Ionization Isomerism

When 0.1 mo...

Question

When $0.1 m o l C o C l_{3} (N H_{3})_{5}$ is treated with excess of $A g N O_{3}, 0.2 m o l$ of $A g C l$ 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 C $1 : 2$ electrolyte
$0.2 m o l$ of $A g C l$ are obtained when $0.1 m o l C o C l_{3} (N H_{3})_{5}$ is treated with excess of $A g N O_{3}$ which shows that one molecule of the complex gives two $C l^{-}$ ions in solution.
Thus, the formula of the complex is $[C o (N H_{3})_{5} C l] C l_{2}$ i.e. $1 : 2$ electrolyte.

Hence, Option "B" is the correct answer.

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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. 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

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}

Q. Resistance of

0.2 M

solution of an electrolyte is

50 o h m

. The specific conductance of the solution is

1.4 S m^{- 1}

. The resistance of

0.5 M

solution of the same electrolyte is

280 o h m

. The molar conductivity of

0.5 M

solution of the electrolyte in

S m^{2} m o l^{- 1}

is:

The ratio of osmotic pressure of a 1:1 electrolyte AB to that of a non electrolyte solute of same concentration is

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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 $0.1 m o l C o C l_{3} (N H_{3})_{5}$ is treated with excess of $A g N O_{3}, 0.2 m o l$ of $A g C l$ are obtained. The conductivity of solution will correspond to: