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Standard XII

Chemistry

Conductance of Electrolytic Solution

When 0.1 ...

Question

When $0.1$ $m o l$ $C o {C l}_{3} {(N H_{3})}_{5}$ is treated with excess of $A g N O_{3},$ $0.2$ mole 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 B $1 : 2$ electrolyte
Formation of $0.2$ mole of $A g C l$ from $0.1$ mole of complex means that there are two ionizable $C l^{-}$ ion.

Hence, formula of the compound is $[C o {(N H_{3})}_{5} C l] {C l}_{2}$ i.e., $1 : 2$ type 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. 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. 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:

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. Resistance of a conductivity cell filled with a solution of an electrolyte of concentration

0.1 M

100 o h m

. The conductivity of this solution is

1.29 S m^{- 1}

. Resistance of the same cell when filled with

0.2 M

of the same solution is

520 o h m

. The molar conductivity of

0.2 M

solution of the electrolyte will be:

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

The correct option is B 1:2 electrolyteFormation of 0.2 mole of AgCl from 0.1 mole of complex means that there are two ionizable Cl− ion. Hence, formula of the compound is [Co(NH3)5Cl]Cl2 i.e., 1:2 type electrolyte.

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

The correct option is B $1 : 2$ electrolyte
Formation of $0.2$ mole of $A g C l$ from $0.1$ mole of complex means that there are two ionizable $C l^{-}$ ion.

Hence, formula of the compound is $[C o {(N H_{3})}_{5} C l] {C l}_{2}$ i.e., $1 : 2$ type electrolyte.