Question

If \[10{\text{mL}}\] of \[{\text{HCl}}\] solution gave \[0.1435{\text{g}}\] of \[{\text{AgCl}}\] when treated with excess of \[{\text{AgN}}{{\text{O}}_3}\] . The normality of \[{\text{HCl}}\] solution is:
A.\[0.1{\text{N}}\]
B.\[0.01{\text{N}}\]
C.\[10{\text{N}}\]
D.\[1{\text{N}}\]

Answer 1

Hint:Stoichiometry is for the determination of quantities of reactant and products involved in balanced chemical reaction. For example: \[{{\text{H}}_2} + {\text{C}}{{\text{l}}_2} \to 2{\text{HCl}}\] , in this one mole of hydrogen reacts with one mole of chlorine to produce 2 mole of hydrochloric acid. Hence \[2{\text{g}}\] of hydrogen reacts with \[71{\text{g}}\] of chlorine to produce \[73{\text{g}}\] of hydrochloric acid.
Formula Used:
\[{\text{normality}} = \dfrac{{{\text{weight}}\left( {{\text{in gram}}} \right)}}{{{\text{equivalent weight}}}} \times \dfrac{{1000}}{{{\text{volume}}\left( {{\text{in mL}}} \right)}}\]

Complete step by step answer:
For a given balanced chemical equation, information of reactant or product can be determined if information of one of the species is given either in terms of moles, molecules, weight or atoms.
The chemical reaction equation of hydrochloric acid with silver nitrate to form precipitate of silver chloride will be as follow: \[{\text{HCl}} + {\text{AgN}}{{\text{O}}_3} \to {\text{AgCl}} + {\text{HN}}{{\text{O}}_3}\].
We can say that when one mole of hydrochloric acid reacts with one mole of silver nitrate, this results in formation of one mole of silver chloride and one mole of nitric acid. As one mole of hydrochloric acid weight \[36.5{\text{g}}\] and one mole of silver chloride weight \[143.5{\text{g}}\]. So, we can say \[36.5{\text{g}}\] of hydrochloric acid is required to produce \[143.5{\text{g}}\] of silver chloride.
 According to the given data in question, the weight of silver chloride formed after the reaction is \[0.1435{\text{g}}\] . Thus, weight of hydrochloric acid required to produce \[0.1435{\text{g}}\] of silver chloride can be given as: \[{\text{Amount of HCl}}\left( {{\text{in gram}}} \right){\text{required}} = \dfrac{{36.5}}{{143.5}} \times 0.1435 = 0.0365{\text{g}}\] .
As given that volume of hydrochloric acid taken to be is\[10{\text{mL}}\]. Thus, normality of hydrochloric acid can be given as:
\[{\text{normality}} = \dfrac{{{\text{weight}}\left( {{\text{in gram}}} \right)}}{{{\text{equivalent weight}}}} \times \dfrac{{1000}}{{{\text{volume}}\left( {{\text{in mL}}} \right)}}\]
Substituting the values:
\[{\text{normality}} = \dfrac{{0.0365}}{{36.5}} \times \dfrac{{1000}}{{10}}\]
\[{\text{normality}} = 0.1{\text{N}}\]

Thus, the correct option is A.

Note:
 The equivalent weight of a substance is the number of parts by weight of the substance that combine with or displace directly or indirectly \[1.008\] parts by weight of hydrogen or 8 parts by weight of oxygen or \[35.5\] parts by weight of chlorine.