Question

$0.0125\mathrm{M}$ of ${\mathrm{AgNO}}_3$ requiring precipitated Chloride ions in $0.3\mathrm{g}$ of$\left[\mathrm{C}{\left({\mathrm{NH}}_3\right)}_6\right]{\mathrm{Cl}}_3$. What is the volume of it (in ml)?

• A. ${\mathrm{M}}_{\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_6\right]{\mathrm{Cl}}_3}=267.46\mathrm{g}/\mathrm{mol}$
• B. ${\mathrm{M}}_{\left({\mathrm{AgNO}}_3\right)}$

Answer 1

The correct option is A

${\mathrm{M}}_{\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_6\right]{\mathrm{Cl}}_3}=267.46\mathrm{g}/\mathrm{mol}$

Explanation for the correct options

(A)${\mathrm{M}}_{\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_6\right]{\mathrm{Cl}}_3}=267.46\mathrm{g}/\mathrm{mol}$

• $$\begin{gathered} \left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_6\right]{\mathrm{Cl}}_3\left(\mathrm{s}\right)+3{\mathrm{AgNO}}_3\left(\mathrm{s}\right)\to \left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_6\right]{\left({\mathrm{NO}}_3\right)}_3\left(\mathrm{s}\right)+{\mathrm{AgCl}}_3\left(\mathrm{s}\right) \\ \mathrm{Hexaammine}\mathrm{Silver}\mathrm{Nitrate}\mathrm{Hexaammine}\mathrm{Silver} \\ \mathrm{cobalt}\left(\mathrm{III}\right)\mathrm{Chloride}\mathrm{cobalt}\left(\mathrm{III}\right)\mathrm{Nitrate}\mathrm{Chloride} \end{gathered}$$
• Expression used for the calculation of the volume:
• $\frac{\mathrm{Weight}}{\mathrm{Molecular}\mathrm{weight}}\times \mathrm{n}\mathrm{factor}=\mathrm{Molarity}\times \mathrm{Volume}$
• putting values in the above expression:
• $\begin{array}{rcl}\frac{0.3}{267.46}\times 3& =& 0.125\times V\times {10}^{-3}\\ V& =& \frac{0.3\times 3\times 1000}{267.46\times 0.125}\\ & =& 26.92\mathrm{mL}\end{array}$

Final Answer: Hence option (A) is correct.,$26.92\mathrm{mL}$.