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Question: A dilute solution of \[{H_2}S{O_4}\] is made by adding \[{\text{5}}\,{\text{mL}}\]of 3N \[{H_2}S{O_4...

A dilute solution of H2SO4{H_2}S{O_4} is made by adding 5mL{\text{5}}\,{\text{mL}}of 3N H2SO4{H_2}S{O_4}to 245mL{\text{245}}\,{\text{mL}} of water. Find the normality and molarity of the diluted solution.

Explanation

Solution

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.

Complete step by step answer:
We know from normality that,
N1V1=N2V2{{\text{N}}_{\text{1}}}{{\text{V}}_{\text{1}}} = {{\text{N}}_{\text{2}}}{{\text{V}}_{\text{2}}}
Here,
N1{{\text{N}}_{\text{1}}}= normality of H2SO4{H_2}S{O_4}
N2{{\text{N}}_{\text{2}}}= normality of water
V1{{\text{V}}_{\text{1}}}= volume of H2SO4{H_2}S{O_4}
V2{{\text{V}}_{\text{2}}}= volume of water
We are given that,
N1{{\text{N}}_{\text{1}}}= 3N
N2{{\text{N}}_{\text{2}}}= x
V1{{\text{V}}_{\text{1}}}= 5mL{\text{5}}\,{\text{mL}}
V2{{\text{V}}_{\text{2}}}=245mL{\text{245}}\,{\text{mL}}
Using these values in the above equation we will calculate the value of N2{{\text{N}}_{\text{2}}}.

N1V1=N2V2 3×5=x×245 x = 0.0612N  {{\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 0.0612N{\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,

NormalityMolarity=Basicity 0.0612Molarity=2 Molarity=0.0306M  \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 0.0306M{\text{0}}{\text{.0306}}\,{\text{M}}.

Note:
The number of the atoms in 1 mole is equal to the Avogadro’s number denoted as (NA)\left( {{{\text{N}}_{\text{A}}}} \right). The value of the Avogadro’s number is 6.022×10236.022 \times {10^{23}}. The molar mass is given as mass divided by mole that is,
Molar mass mass/moles=g/molesmass/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 12grams{\text{12}}\,{\text{grams}} per mole.