When 0.01 mole of a cobalt complex is treated with excess silver nitrate solution, 4.305 g silver chloride is precipitated. The formula of the complex is:

[C o (N H_{3})_{6}] C l_{3}

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[C o (N H_{3})_{6} C l_{3}]

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

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[C o (N H_{3})_{4} C l_{2}] N O_{3}

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Solution

The correct option is D $[C o (N H_{3})_{6}] C l_{3}$
$4.305 g A g C l = \frac{4.305}{143.5} m o l = 0.03 m o l$
As 0.01 mole of the complex gives 0.03 mole of $A g C I$ , this shows that there are three ionisable $C l$ .

Hence, formula is $[C o (N H_{3})_{6}] C l_{3}$ .

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Similar questions

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. 0.02 mole of

[C o (N H_{3})_{5} B r] C l_{2}

and 0.02 mole of

[C o (N H_{3})_{5} C l] S O_{4}

are present in 200 cc of a solution X. The number of moles of the preceipitates Y and Z that are formed when the solution X is treated with excess silver nitrate and excess barium chloride are respectively :

Q. The number of chloride ions which would be precipitated, when

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

is treated with silver nitrate solution:

Q. The correct order of the following complexes according to increasing conductivity of their solutions will be:
(i)

[C o (N H_{3})_{3} C l_{3}]

(ii)

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

(iii)

[C o (N H_{3})_{6}] C l_{3}

(iv)

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

Q. 2.66 g chloride of a metal when treated with silver nitrate solution give 2.87 g of silver chloride. 3.37 g of another chloride of the same metal give 5.74 g of silver chloride when treated with silver nitrate solution. Show that the results are in agreement with a law of chemical combination.

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When 0.01 mole of a cobalt complex is treated with excess silver nitrate solution, 4.305 g silver chloride is precipitated. The formula of the complex is:

The correct option is D [Co(NH3)6]Cl34.305gAgCl=4.305143.5mol=0.03molAs 0.01 mole of the complex gives 0.03 mole of AgCI, this shows that there are three ionisable Cl. Hence, formula is [Co(NH3)6]Cl3.

The correct option is D $[C o (N H_{3})_{6}] C l_{3}$
$4.305 g A g C l = \frac{4.305}{143.5} m o l = 0.03 m o l$
As 0.01 mole of the complex gives 0.03 mole of $A g C I$ , this shows that there are three ionisable $C l$ .

Hence, formula is $[C o (N H_{3})_{6}] C l_{3}$ .