Question

Molarity of $\text{0}\text{.2N}\,\,{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}$ is:
(A) $0.2$
(B) $0.4$
(C) $0.6$
(D) $0.1$

Answer 1

Hint: Molarity of a solution is defined as the number of solute particles present in $1\text{L}$ of solution.
\[\text{Molarity(M) = }\dfrac{\text{moles}\,\text{of solute}}{\text{Volume of}\,\text{solution}\,(L)}\]
Normality is the number of grams equivalent of solute present in $1\text{L}$ of solution.
 \[\text{Normality(N) }\,\text{= }\dfrac{\text{No of}\,\text{gm}\,\text{equivalent of solute}}{\text{Volume}\,\text{of}\,\text{solution}\,\text{(L)}}\]
Relation in between normality and molarity
$\begin{align}
  & \text{N =}\,\text{M}\,\text{ }\!\!\times\!\!\text{ f}......\text{(i)} \\
 & \text{where,}\,\,\,\text{f =}\,\,\text{valency}\,\text{factor} \\
\end{align}$

Complete answer:
Valency factor (f) is defined as the ability of an atom to lose or gain electrons. So their acidity and basicity is the valency factor for a base and acid respectively.
\[\begin{align}
  & ~~~~~\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ Normality}\,\text{of}\,\text{acid = }\!\!~\!\!\text{ Molarity}\,\,\text{ }\!\!\times\!\!\text{ }\,\,\text{basicity} \\
 & \,\,\,\,\,\,\,\,\,\,\text{Normality}\,\text{of}\,\text{base =}\,\text{Molarity}\,\,\text{ }\!\!\times\!\!\text{ }\,\,\text{acidity} \\
\end{align}\]
By applying the equation (i) we will calculate the molarity of the solution.
Relation in between normality and molarity
$\begin{align}
  & \text{N=}\,\text{M}\,\text{ }\!\!\times\!\!\text{ f} \\
 & \text{M=}\,\dfrac{\text{N}}{\text{f}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{ }\!\!\{\!\!\text{ }\,\,\text{f }\,\text{= 2 }\,\text{for}\,{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\text{ }\!\!\}\!\!\text{ } \\
 & \text{M=}\dfrac{\text{0}\text{.2}}{\text{2}}\text{=}\,\text{0}\text{.1} \\
\end{align}$

So, option (D) is the correct answer.

Additional information:
Molar solution means one mole of solute is present in the one litter of solution. In molarity and normality the volume of the solution is considered, while in molality mass of the solvent is considered.
Normality and molarity changes with temperature because of expansion or contraction of the liquid with temperature. However, the molality of a solution does not change with temperature because the mass of the solvent does not change with temperature.
Normality can be equal and greater than molarity but cannot be smaller than molarity of the solution.
In redox reaction change in oxidation number will determine the valency factor of the molecule.

Note:
\[\begin{align}
& \text{N = }\dfrac{\text{No of gram eq}\text{. of solute}}{\text{1L}}.......(i) \\
& \text{No of gram eq}\text{. of solute }\,\text{=}\,\,\dfrac{\text{Weight(W)}}{\text{eq}\text{.}\,\,\text{weight}}=\dfrac{\text{Weight(W)}}{{}^{\text{molar}\,\text{weight}}/{}_{\text{factor}}} \\
 & \\
\end{align}\]
So, from equation (i) and equation (ii) we get
\[\begin{align}
  & \text{0}\text{.2=}\dfrac{\text{W}}{{}^{\text{98}}/{}_{\text{2}}} \\
 & \text{W=}\,\,\text{49 }\!\!\times\!\!\text{ 0}\text{.2} \\
 & \text{W=}\,\,\text{9}\text{.8} \\
\end{align}\]
So, no. of moles of ${{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}$ \[\text{mole(n) =}\,\dfrac{\text{W}}{\text{M}\text{.W}}\text{ = }\dfrac{\text{9}\text{.8}}{\text{98}}\text{ = 0}\,\text{.1}\]
So, molarity of the solution $\text{M=}\dfrac{\text{0}\text{.1}}{\text{1}}\text{=}\,\text{0}\text{.1}$