Question
The ratio of the number of moles of $$AgN{O_3},$$ $$Pb{\left( {N{O_3}} \right)_2}$$ and $$Fe{\left( {N{O_3}} \right)_3}$$ required for coagulation of a definite amount of a colloidal sol of silver iodide prepared by mixing $$AgN{O_3}$$ with excess of $$KI$$ will be
A.
1 : 2 : 3
B.
3 : 2 : 1
C.
6 : 3 : 2
D.
2 : 3 : 6
Answer :
6 : 3 : 2
Solution :
With excess of $$KI,$$ colloidal particles will be $$\left[ {AgI} \right]{I^ - }$$
\[\underset{1\,\,mol}{\mathop{\left[ AgI \right]{{I}^{-}}}}\,+\underset{1\,\,mol}{\mathop{AgN{{O}_{3}}}}\,\to AgI\downarrow +\] \[AgI\downarrow +NO_{3}^{-}\]
\[\underset{\begin{smallmatrix}
2\,\,mol \\
1\,\,mol
\end{smallmatrix}}{\mathop{2\left[ AgI \right]{{I}^{-}}}}\,+\underset{\begin{smallmatrix}
1\,\,mol \\
\frac{1}{2}\,mol
\end{smallmatrix}}{\mathop{Pb{{\left( N{{O}_{3}} \right)}_{2}}}}\,\to \] \[2AgI\downarrow +Pb{{I}_{2}}\downarrow +2NO_{3}^{-}\]
\[\underset{\begin{smallmatrix}
3\,\,mol \\
1\,\,mol
\end{smallmatrix}}{\mathop{3\left[ AgI \right]{{I}^{-}}}}\,+\underset{\begin{smallmatrix}
1\,\,mol \\
\frac{1}{3}\,mol
\end{smallmatrix}}{\mathop{Fe{{\left( N{{O}_{3}} \right)}_{3}}}}\,\to \] \[3AgI\downarrow +Fe{{I}_{3}}\downarrow +3NO_{3}^{-}\]
∴ Molar ratio required for coagulation of same amount of $$\left[ {AgI} \right]{I^ - }\,\,{\text{is}} = 1:\frac{1}{2}:\frac{1}{3} = 6:3:2$$