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Conductance of Electrolytic Solution

A complex P...

Question

A complex $P t {C l}_{4} .5 {N H}_{3}$ shows a molar conductance of $402 {o h m}^{- 1}$ ${c m}^{- 1}$ ${m o l}^{- 1}$ in water and precipitates three moles of $A g C l$ with $A g {N O}_{3}$ solution. The formula of the complex is :

A

[P t {({N H}_{3})}_{6}] {C l}_{4}

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B

[P t {({N H}_{3})}_{6} {C l}_{2}] C l_{2}

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C

[P t {({N H}_{3})}_{5} C l] {C l}_{3}

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D

[P t {({N H}_{3})}_{3} {C l}_{3}] {C l}

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Solution

The correct option is B $[P t {({N H}_{3})}_{5} C l] {C l}_{3}$
The complex $P t {C l}_{4} .5 {N H}_{3}$ is designated as $[P t {({N H}_{3})}_{5} C l] {C l}_{3}$ which ionizes to ${[P t {({N H}_{3})}_{5} C l]}^{3 +}$ $+ 3 {C l}^{-}$ ions.

Thus total ions produced are $f o u r$ but $t h r e e$ moles of $A g C l$ are produced from $3 {C l}^{-}$ ions with $A g {N O}_{3}$ .

Hence, option $C$ is correct.

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Similar questions

P t C l_{4} .5 N H_{3}

shows a molar conductance of

402 {ohm}^{- 1} {cm}^{2} {mol}^{- 1}

in water and precipitates 3 moles of

A g C l

A g N O_{3}

solution. The formula of the complex is:

P t C l_{4} .5 N H_{3}

shows a molar conductance of

402 {ohm}^{- 1} {cm}^{2} {mol}^{- 1}

in water and precipitates 3 moles of

A g C l

A g N O_{3}

solution. The formula of the complex is:

Q. When 1 mole of

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

is treated with excess of

A g N O_{3}

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

Arrange the following complexes in the increasing order of conductivity of their solution

$[C o (N H_{3})_{3} C l_{3}], [C o (N H_{3})_{4} C l_{2}] C l, [C o (N H_{3})_{6}] C l_{3}, [C o (N H_{3})_{5} C l] C l_{2}$

Q. Combination of

P t (I V), N H_{3}, C l^{-}, K^{+}

result in the formation of

x

complexes and one such complex is

[P t (N H_{3})_{6}] C l_{4}

x

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