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Question: An excess of \[AgN{O_3}\] is added to \[100ml\] of a \[0.01M\] solution of Dichlorotetraaquachromium...

An excess of AgNO3AgN{O_3} is added to 100ml100ml of a 0.01M0.01M solution of Dichlorotetraaquachromium(111)(111)chloride. The number of moles of AgClAgCl precipitate would be:
(A) 0.0010.001
(B) 0.0020.002
(C) 0.0030.003
(D) 0.010.01

Explanation

Solution

Dichlorotetraaquachromium (111)(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)(111)chloride is 100ml=100/1000=0.1L100ml = 100/1000 = 0.1L
Concentration of Dichlorotetraaquachromium(111)(111) chloride is 0.01M0.01M.
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)(111) its molecular formula represented as [Cr(H20)4Cl2]+Cl1{\left[ {Cr{{({H_2}0)}_4}C{l_2}} \right]^ + }C{l^{ - 1}} so it is an example heteroleptic compound.
When 1mol1mol of [Cr(H20)4Cl2]+Cl1{\left[ {Cr{{({H_2}0)}_4}C{l_2}} \right]^ + }C{l^{ - 1}} react with excess of AgNO3AgN{O_3}then 1mol1mol of AgClAgCl is formed because one chloride ion is present for replacement. So we first calculate how much mole is Dichlorotetraaquachromium(111)(111) in 0.01M0.01M having volume 100ml=100/1000=0.1L100ml = 100/1000 = 0.1L.
Number of moles of Dichlorotetraaquachromium(111)(111)chloride = Volume×molar concentrationVolume \times {\text{molar concentration}}
Number of moles of Dichlorotetraaquachromium(111)(111)chloride =0.1×0.01M = 0.1 \times 0.01M
Number of moles of Dichlorotetraaquachromium(111)(111)chloride=0.001mol0.001mol
When 1mol1mol of [Cr(H20)4Cl2]+Cl1{\left[ {Cr{{({H_2}0)}_4}C{l_2}} \right]^ + }C{l^{ - 1}} react with excess of AgNO3AgN{O_3}then 1mol1mol of AgClAgCl is formed
So When 0.001mol0.001mol of [Cr(H20)4Cl2]+Cl1{\left[ {Cr{{({H_2}0)}_4}C{l_2}} \right]^ + }C{l^{ - 1}} react with excess of AgNO3AgN{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.