Question

An excess of \[AgN{O_3}\] is added to \[100ml\] of a \[0.01M\] solution of Dichlorotetraaquachromium\[(111)\]chloride. The number of moles of \[AgCl\] precipitate would be:
(A) \[0.001\]
(B) \[0.002\]
(C) \[0.003\]
(D) \[0.01\]

Answer 1

Hint:Dichlorotetraaquachromium \[(111)\] chloride it is a coordination compound and coordination compound is formed when a central metal atom which is positively charged is attached to the negatively charged or neutral electrons rich ligand to form coordination compound.

Complete step by step answer:
Given- volume of Dichlorotetraaquachromium\[(111)\]chloride is \[100ml = 100/1000 = 0.1L\]
Concentration of Dichlorotetraaquachromium\[(111)\] chloride is \[0.01M\].
Coordination compounds are two types: Homoleptic coordination compound- A complex having one type of ligands in coordination compound that compound is known as homoleptic coordination compound.
Heteroleptic coordination compound- A complex having more than type of ligands in a coordination compound that is known as Heteroleptic coordination compound.
Dichlorotetraaquachromium\[(111)\] its molecular formula represented as \[{\left[ {Cr{{({H_2}0)}_4}C{l_2}} \right]^ + }C{l^{ - 1}}\] so it is an example heteroleptic compound.
When \[1mol\] of \[{\left[ {Cr{{({H_2}0)}_4}C{l_2}} \right]^ + }C{l^{ - 1}}\] react with excess of \[AgN{O_3}\]then \[1mol\] of \[AgCl\] is formed because one chloride ion is present for replacement. So we first calculate how much mole is Dichlorotetraaquachromium\[(111)\] in \[0.01M\] having volume \[100ml = 100/1000 = 0.1L\].
Number of moles of Dichlorotetraaquachromium\[(111)\]chloride = \[Volume \times {\text{molar concentration}}\]
Number of moles of Dichlorotetraaquachromium\[(111)\]chloride \[ = 0.1 \times 0.01M\]
Number of moles of Dichlorotetraaquachromium\[(111)\]chloride=\[0.001mol\]
When \[1mol\] of \[{\left[ {Cr{{({H_2}0)}_4}C{l_2}} \right]^ + }C{l^{ - 1}}\] react with excess of \[AgN{O_3}\]then \[1mol\] of \[AgCl\] is formed
 So When \[0.001mol\] of \[{\left[ {Cr{{({H_2}0)}_4}C{l_2}} \right]^ + }C{l^{ - 1}}\] react with excess of \[AgN{O_3}\]then of \[0.001mol\]\[AgCl\] is formed.
Hence option A is the correct option.

Note:
In coordination compound metal atoms or ions and its ligands are always placed in a square bracket that square bracket is known as a coordination entity.
And the total number of ligands around the present central metal atom or ion is known as coordination number. Coordination number is always equal to the number of unidentate ligands or double to the bidentate ligands.
Coordination compounds are stable at solid state as well as in aqueous state. In aqueous medium it forms simple ions but coordination compounds maintain identity itself in aqueous medium.