Question

# An excess of $AgN{{O}_{3}}$ is added to 100ml of a 0.01M solution of dichlorotetraaquachromium(III)chloride. The number of moles of AgCl precipitate would be:A. 0.001B. 0.002C. 0.003D.0.01

Hint: $\left[ Cr{{\left( {{H}_{2}}O \right)}_{4}}C{{l}_{2}} \right]$ is a coordination compound. Here the overall charge on the compound is +1. We can calculate the number of moles by the formula:$Number\text{ }of\text{ }moles=Volume\times Molarity$
Formula used: $Number\text{ }of\text{ }moles=Volume\times Molarity$
- We can write the formula of given compound as $\left[ Cr{{\left( {{H}_{2}}O \right)}_{4}}C{{l}_{2}} \right]$. Now, we will first find the charge on the compound. We know that charge on chromium is +3, charge on water is zero, charge on chlorine is -1. So, we get the overall charge on the compound as:
\begin{align} & ch\arg e=3+0\times \left( 4 \right)+2\times \left( -1 \right) \\ & =+1 \\ \end{align}
So, we can see that the overall charge present is +1. Therefore, there will be one chloride ion required.
- We can say that if one mole of a complex is present then one mole chloride ion will be available to form AgCl. In simple words we can say that 1 mole of complex will give 1 mole of AgCl.
- Now let’s calculate the number of moles, we can use the formula:
$Number\text{ }of\text{ }moles = Volume\times Molarity$
-We are being provided with the volume and molarity, we should first convert the volume given in ml into litres. So,100 ml = 0.1 l
-So, we will put the values in the equation,
\begin{align} & Number\text{ }of\text{ }moles=0.1\times 0.01 \\ & =0.001\text{ }moles \\ \end{align}
Hence, we can conclude that the correct option is (A), that is the number of moles of AgCl precipitated would be 0.001 moles.

Note: Here, we are being provided with a 0.01 M solution, that is M is the molarity of the solution. One should not be confused in units’ M and m. M is the unit of molarity and m is the unit of molality.