Question

Excess silver nitrate solution is added to $100ml$ of $0.01{\text{ }}M$ pentaaquachlorochromium $\left( 3 \right)$ chloride solution. The mass of the silver chloride obtained in grams is:
A.$287 \times {10^{ - 3}}$
B.$143.5 \times {10^{ - 3}}$
C.$143.5 \times {10^{ - 3}}$
D.$2.87 \times {10^{ - 2}}$

Answer 1

The answer to this question can only be obtained after we write the compound pentaaquachlorochromium $\left (3 \right)$ chloride in its molecular form. Therefore, the molecular form of this solution is $\left[ {Cr{{\left( {{H_2}O} \right)}_5}Cl} \right]C{l_2}$ . This is the reagent that will release $C{l^ - }$ ions.
FORMULA USED: $Molarity\left( M \right) = \dfrac{n}{V}$
Where $n$ is the number of moles and $V$ is the volume of solution in liters.
$n = \dfrac{\text{given weight}} {\text{molecular weight}}$

Complete step by step answer:
The solution made of $\left[ {Cr{{\left( {{H_2}O} \right)}_5}Cl} \right]C{l_2}$ undergoes dissociation to release chloride ions. To be precise, by looking at the compound we come to know that $2$ $C {l^ - }$ ions are released upon dissociation. In addition to this we are also told that an excess of silver nitrate solution that is, $AgN{O_3}$ is added into the solution.
The addition of $AgN{O_3}$ will lead to the formation of silver chloride that is, $AgCl$ as $AgN{O_3}$ will combine with the free chloride ions. We are now asked to measure the mass of $AgCl$ that is found in the solution. The answer will be derived in the following steps.
Calculate the number of moles of chloride ions present in the solution. For this we employ the use of the information about the number of moles of $\left [ {Cr {{\left({{H_2} O} \right)} _5} Cl} \right]C{l_2}$ present in the solution. According to the question, $100ml$ of $0.01{\text{}}M$ of $\left [ {Cr {{\left({{H_2} O} \right)} _5} Cl} \right]C{l_2}$ is used. And each molecule of $\left [ {Cr {{\left({{H_2} O} \right)} _5} Cl} \right]C{l_2}$ releases two chloride ions. Therefore, using the volume and molarity of the solution given, we can derive the number of ions present as shown below:
$Molarity\left( M \right) = \dfrac{n}{V}$
$\therefore n = M \times V$
On substituting the values we get ,
$\Rightarrow n = \dfrac{{2 \times 0.01 \times 100}}{{1000}}\left( l \right)$
Here we have divided by $1000$ , because the volume has been given in milliliters.
Now we can find the number of moles of chloride ions formed to be,
$n = 2 \times 0.001$
$\Rightarrow n = 0.002$
Calculate the number of moles of $AgCl$ formed. Now each chloride ion binds with one silver ion. Since we know that there are $0.002$ moles or chloride ions, we can infer that the number of moles of $AgCl$ will also be $0.002$.
Mass of $AgCl$: to find mass we have to just multiply the mass of one $AgCl$ molecule with the number of moles. This can be done as follows:
$Mass{\text{ }}ofAgCl = {\text{ }}mass{\text{ }}of{\text{ }}A{g^ + } + mass{\text{ }}of{\text{ C}}{{\text{l}}^ - }$
On substituting the values we get ,
$\Rightarrow 108 + 35.5$
$\Rightarrow 143.5$
Therefore, total mass of $AgCl$ found is,
$Mass = 143.5 \times 0.002$
$\Rightarrow 0.287$
$\Rightarrow 287 \times {10^ {- 3}}$
Therefore, the answer will be option A.

Note: The number of moles of a solution can also be found by multiplying the volume of the solution with the number of moles.
complexes have the ability to break up into their constituent ions. This is seen in the atoms which are not directly attached to the central metal atom but to the coordination sphere as seen in the case of the chlorine ions in the complex given above.