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Question: Assuming that \[{\text{100 cc}}{\text{.}}\] of \[{\text{0}}{\text{.1M}}\] the solution \[{H_2}S{O_4}...

Assuming that 100 cc.{\text{100 cc}}{\text{.}} of 0.1M{\text{0}}{\text{.1M}} the solution H2SO4{H_2}S{O_4} is needed to neutralize 200 cc{\text{200 cc}} of solution the normality of Ba(OH)2Ba{\left( {OH} \right)_2} a solution is:
A.0.01N{\text{0}}{\text{.01N}}
B.0.5N{\text{0}}{\text{.5N}}
C.0.1N{\text{0}}{\text{.1N}}
D.0.05N{\text{0}}{\text{.05N}}

Explanation

Solution

To answer this question, you should recall the concept of normality and neutralization reaction. When equal equivalents of an acid and a base are mixed it leads to a neutral solution.
The formula used:
Normality = Molarity×n - factor{\text{Normality = Molarity}} \times {\text{n - factor}}

Complete step by step answer:
In the question, it is mentioned that Molarity of H2SO4{H_2}S{O_4}  = 0.1 M{\text{ = 0}}{\text{.1 M}}.
n-factor of H2SO4{H_2}S{O_4} = 2 (As seen from the formula that it has 2 dissociableH{\text{H}}).
So, Normality of H2SO4{H_2}S{O_4} = 0.1×2 N = 0.2 N = {\text{ }}0.1 \times 2{\text{ }}N{\text{ }} = {\text{ }}0.2{\text{ }}N
At equivalence point Gram equivalents of H2SO4{H_2}S{O_4} = Gram equivalents of Ba(OH)2Ba{\left( {OH} \right)_2}
So we can conclude that:
NormalityH2SO4×VolumeH2SO4=NormalityBa(OH)2×VolumeBa(OH)2{\text{Normalit}}{{\text{y}}_{{H_2}S{O_4}}} \times {\text{Volum}}{{\text{e}}_{{H_2}S{O_4}}} = {\text{Normalit}}{{\text{y}}_{Ba{{\left( {OH} \right)}_2}}} \times {\text{Volum}}{{\text{e}}_{Ba{{\left( {OH} \right)}_2}}}.
Given that:
NormalityH2SO4=0.2N{\text{Normalit}}{{\text{y}}_{{H_2}S{O_4}}} = 0.2{\text{N}}, VolumeH2SO4=100cc{\text{Volum}}{{\text{e}}_{{H_2}S{O_4}}} = 100{\text{cc}}and VolumeBa(OH)2=200cc{\text{Volum}}{{\text{e}}_{Ba{{\left( {OH} \right)}_2}}} = 200{\text{cc}}.
Substituting these values in the above equation we get:
⇒0.2×100=NormalityBa(OH)2×200\Rightarrow 0.2 \times 100 = {\text{Normalit}}{{\text{y}}_{Ba{{\left( {OH} \right)}_2}}} \times 200.
We will get NormalityBa(OH)2=0.1N{\text{Normalit}}{{\text{y}}_{Ba{{\left( {OH} \right)}_2}}} = 0.1{\text{N}}.

Hence, the correct answer to this question is option C.

Note:
You should know about the other concentration terms commonly used:
Concentration in Parts Per Million (ppm) The parts of a component per million parts (106{10^6}) of the solution.
ppm(A)=Mass of ATotal mass of the solution×106{\text{ppm(A)}} = \dfrac{{{\text{Mass of A}}}}{{{\text{Total mass of the solution}}}} \times {10^6}
Molality (m): Molality establishes a relationship between moles of solute and the mass of solvent. It is given by moles of solute dissolved per kg of the solvent. The molality formula is as given- Molality(m) = Mole of soluteMass of solvent in kg{\text{Molality(m) = }}\dfrac{{{\text{Mole of solute}}}}{{{\text{Mass of solvent in kg}}}}