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# A dilute solution of ${H_2}S{O_4}$ is made by adding ${\text{5}}\,{\text{mL}}$of 3N ${H_2}S{O_4}$to ${\text{245}}\,{\text{mL}}$ of water. Find the normality and molarity of the diluted solution.

Last updated date: 09th Aug 2024
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Hint: Molality or molal concentration, is defined as the amount of the substance which is dissolved in a certain mass of the solvent. It is also defined as the moles of a solute per kilogram of a solvent. Normality is defined as the number of the mole or gram equivalents of solute present in one litre of a solution.

We know from normality that,
${{\text{N}}_{\text{1}}}{{\text{V}}_{\text{1}}} = {{\text{N}}_{\text{2}}}{{\text{V}}_{\text{2}}}$
Here,
${{\text{N}}_{\text{1}}}$= normality of ${H_2}S{O_4}$
${{\text{N}}_{\text{2}}}$= normality of water
${{\text{V}}_{\text{1}}}$= volume of ${H_2}S{O_4}$
${{\text{V}}_{\text{2}}}$= volume of water
We are given that,
${{\text{N}}_{\text{1}}}$= 3N
${{\text{N}}_{\text{2}}}$= x
${{\text{V}}_{\text{1}}}$= ${\text{5}}\,{\text{mL}}$
${{\text{V}}_{\text{2}}}$=${\text{245}}\,{\text{mL}}$
Using these values in the above equation we will calculate the value of ${{\text{N}}_{\text{2}}}$.
${{\text{N}}_{\text{1}}}{{\text{V}}_{\text{1}}} = {{\text{N}}_{\text{2}}}{{\text{V}}_{\text{2}}} \\ \Rightarrow 3 \times 5\, = \,{\text{x}} \times \,{\text{245}} \\ \Rightarrow \,{\text{x = }}\,{\text{0}}{\text{.0612}}\,{\text{N}} \\$
Therefore, the value of normality is ${\text{0}}{\text{.0612}}\,{\text{N}}$.
Now, we will calculate the value of molarity.
We know that the basicity of sulphuric acid is = 2.
The relation between normality, molarity and basicity is given as,
$\dfrac{{{\text{Normality}}}}{{{\text{Molarity}}}} = {\text{Basicity}} \\ \Rightarrow \dfrac{{{\text{0}}{\text{.0612}}}}{{{\text{Molarity}}}} = {\text{2}} \\ \Rightarrow {\text{Molarity}} = {\text{0}}{\text{.0306}}\,{\text{M}} \\$
$\therefore$, the value of molarity of the diluted solution is ${\text{0}}{\text{.0306}}\,{\text{M}}$.

Note:
The number of the atoms in 1 mole is equal to the Avogadro’s number denoted as $\left( {{{\text{N}}_{\text{A}}}} \right)$. The value of the Avogadro’s number is $6.022 \times {10^{23}}$. The molar mass is given as mass divided by mole that is,
Molar mass $mass/moles=g/moles$.
The mole concept also concludes that the atomic mass of the Carbon-12 is equivalent to the 12 atomic mass units. Also, the mass of Carbon-12 is exactly ${\text{12}}\,{\text{grams}}$ per mole.