Question

In the complex with formula $MC{l_3}.4{H_2}O$, the coordination number of the metal M is six and there is no molecule of hydration in it. The volume, of $0.1M$ $AgN{O_3}$ solution needed to precipitate the free chloride ions in $200mL$ of $0.01M$ solution of the complex is:
(a)$40mL$
(b)$20mL$
(c)$60mL$
(d)$80mL$

Answer 1

Hint: The coordination number is defined as the number of atoms, molecules and ions bonded to a central atom in a molecule. Coordination number is also known as ligancy of a central atom in a crystal. The ion or molecule which surrounds the central atom or molecule is known as a ligand.

Complete answer:
We can see that the coordination number of the metal M is six and the number of molecules of hydration is zero as already mentioned in the question. Therefore, the formula of the complex is
$ = \left[ {M{{\left( {{H_2}O} \right)}_4}C{l_2}} \right]Cl$
The reaction of the complex with silver nitrate is given below.
$\left[ {M{{\left( {{H_2}O} \right)}_4}C{l_2}} \right]Cl + AgN{O_3} \to \left[ {M{{\left( {{H_2}O} \right)}_4}C{l_2}} \right]N{O_3} + AgCl$
$C{l^ - } + AgN{O_3} \to NO_3^ - + AgCl$
Let the volume of silver nitrate $\left( {AgN{O_3}} \right)$ used $ = V$
Now by using the conservation formula that is,
${M_1}V{C_1}\left( {AgN{O_3}} \right) = {M_2}V{C_2}\left( {C{l^ - }} \right)$
Now putting the required values we get,
$ = 0.1 \times V = 200 \times 0.01$
By solving we get, $V = 20mL$
Therefore, option B is the correct option.

Note:
Silver nitrate is an inorganic compound. It is very less sensitive to light as compared to halides. It is also known as lunar caustic because silver was early known as luna by the ancient alchemists. Silver nitrate is used as a precursor to other silver compounds which are used in photography. It is also non-hygroscopic in comparison with silver fluoroborate and silver perchlorate. It is also soluble in water and other solvents also.