Question

If $\text{ 0}\text{.5 }$ a mole of silver salt is taken and the weight of residue obtained is 216 g. ($\text{ Ag = 108 g / mol }$ ). Then which of the following is correct?
$\text{ }{{\left( \begin{matrix}
 {{\text{C}}_{\text{2}}}{{\text{H}}_{\text{3}}} & {} & {} & {} & {} \\
\text{ }\!\!|\!\!\text{ } & {} & {} & {} & {} \\
\text{C} & - & \text{H} & - & \text{COOH} \\
\text{ }\!\!|\!\!\text{ } & {} & {} & {} & {} \\
{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{3}}} & {} & {} & {} & {} \\
\end{matrix} \right)}_{3}}\text{ + AgN}{{\text{O}}_{\text{3}}}\left( \text{Excess} \right)\to \text{Silver salt}\to \text{Ag(metal) }$

Answer 1

Hint: The silver salt method is used to determine the molar mass of organic acid. A known amount of organic acid (carboxylic acid) is treated with ammonium hydroxide. The ammonium salt is then treated with a silver nitrate solution to obtain silver salt of organic acid. Obtained silver salt is further ignited to obtain the silver metal as the residue.
$\text{ RCOOH }\xrightarrow[\text{AgN}{{\text{O}}_{\text{3}}}]{\text{N}{{\text{H}}_{\text{4}}}\text{OH}}\text{RCOOAg}\xrightarrow{\text{Ignition}}\text{Ag }$

Complete step by step answer:
 We are given the following data:
The number of moles of silver salt is $\text{ n = 0}\text{.5 }$
Weight of residue obtained is equal to $\text{ 216 g }$
The molecular weight of silver is $\text{ 108 g / mol }$
The reaction between the organic molecules and the silver salt is given as follows,
$\text{ }{{\left( {{\left( {{\text{C}}_{\text{2}}}{{\text{H}}_{\text{3}}} \right)}_{2}}\text{C}-\text{H}-\text{COOH} \right)}_{3}}\text{ + AgN}{{\text{O}}_{\text{3}}}\left( \text{Excess} \right)\to \text{Silver salt }\to \text{Ag(metal) }$
When the known mass organic compound is treated with silver nitrate solution, a silver salt of the organic compound is obtained as the production ignition of silver salt we obtain metallic silver as the residue.
Here 216 g of the silver metal residue is obtained from the organic carboxylic acid. One mole of silver metal corresponds to the 108 g of the silver metal. Thus 216 g of the silver metal residue corresponds to the 2 moles of silver metal.
$\begin{align}
& \text{ 1 mole of silver metal = 108 g } \\
& \therefore 216\text{ g of Ag = }\dfrac{216\text{ g }}{\text{108 g}}=\text{ 2 mole } \\
\end{align}$
In the silver salt method, each proton in the carboxylic acid is replaced by the silver ion. This reaction is as shown below,
$\text{ RCOOH }\xrightarrow[\text{AgN}{{\text{O}}_{\text{3}}}]{\text{N}{{\text{H}}_{\text{4}}}\text{OH}}\text{RCOOAg}\xrightarrow{\text{Ignition}}\text{Ag }$
One mole of carboxylic acid corresponds to one mole of silver residue. But our reaction produces $\text{0}\text{.5 }$moles of the silver salt. Thus there are 4 silver atoms in the reaction.
$\begin{align}
& \text{ 2 moles of Ag = n }\times \text{ 0}\text{.5 mole of Ag salt} \\
& \Rightarrow \text{n = }\dfrac{2}{0.5}=\text{ 4 Ag atoms } \\
\end{align}$
Each silver atom has a valency of 1. The total mass of silver metal residue is equal to the product of valency and the atomic mass of silver. It is given as follows,
$\begin{align}
& \text{ 216 = 108 }\times \text{ x }\times \text{ 0}\text{.5 mol} \\
& \Rightarrow \text{x = 4 } \\
\end{align}$
Thus the total silver atoms have valency equal to 4.

Hence, (A) is the correct option.

Note: Note that the silver salt method can be used to determine the molar mass of acid. The formula is stated as follows,
$\text{ Molar mass of acid = }\left[ \dfrac{\text{weight of silver salt}}{\text{weight of silver}}\times 108-107 \right]\times \text{Basicity of acid }$
The basicity of acid is equal to the exchangeable protons of acid.