(a) What is the normality of a 96% solution of ${{H}_{2}}S{{O}_{4}}$ of specific gravity 1.84?
(b) How many mL of 96% sulphuric acid solution is necessary to prepare one litre of 0.1 N ${{H}_{2}}S{{O}_{4}}$?
(c) To what volume should 10 mL of 96% ${{H}_{2}}S{{O}_{4}}$ be diluted to prepare 2 N solution?

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Hint: Normality (N) is defined as the number of gram equivalents of a solute present per litre of the solution. It is given as:
     \[\text{Normality}=\dfrac{\text{weight of solute (in grams)}}{\text{equivalent weight}}\times \dfrac{1000}{\text{volume of solution (in mL)}}\]
if normality of one solution is given, that of other can be calculated using normality equation which is given as
Specific gravity is the ratio of the density of the given substance to the density of water. density of water is 1$gm{{L}^{-1}}$.

Complete answer:
(a) To find the normality of a 96% solution of sulphuric acid, ${{H}_{2}}S{{O}_{4}}$.
- A 96% solution means that 96 grams of ${{H}_{2}}S{{O}_{4}}$ is dissolved in 100 grams of solution.
Given weight of ${{H}_{2}}S{{O}_{4}}$ = 96 grams
Molar mass of ${{H}_{2}}S{{O}_{4}}$ = 98 grams.
Basicity of ${{H}_{2}}S{{O}_{4}}$ is 2. Therefore, equivalent weight will be equal to
  & \text{Equivalent weight = }\dfrac{\text{Molar mass}}{\text{Basicity}}=\dfrac{98g}{2}=49g\,e{{q}^{-1}} \\
 & \\
Given specific gravity of ${{H}_{2}}S{{O}_{4}}$= 1.84. specific gravity is given as
  & \text{Specific gravity = }\dfrac{\text{density of }{{\text{H}}_{2}}S{{O}_{4}}}{\text{density of }{{\text{H}}_{2}}O} \\
 & \text{density of }{{\text{H}}_{2}}S{{O}_{4}}=\text{Specific gravity }\times \text{density of }{{\text{H}}_{2}}O \\
 & \text{density of }{{\text{H}}_{2}}S{{O}_{4}}=1.84\times 1=1.84g\,m{{L}^{-1}} \\
Density of ${{H}_{2}}S{{O}_{4}}$ in the 100 gram solution =$\dfrac{100}{VmL}$
Volume, V of the solution will be = $\dfrac{100g}{1.84gm{{L}^{-1}}}=54.35mL$
Hence, the normality of 96% ${{H}_{2}}S{{O}_{4}}$ with specific gravity 1.84 is calculated to be
  & \text{Normality}=\dfrac{\text{weight of solute (in grams)}}{\text{equivalent weight}}\times \dfrac{1000}{\text{volume of solution (in mL)}} \\
 & N=\dfrac{96g}{49g\,e{{q}^{-1}}}\times \dfrac{1000}{54.35mL}=36.048N \\

(b) - Normality of 96% ${{H}_{2}}S{{O}_{4}}$, ${{N}_{1}}$= 36.048
Let the volume of 96% solution of ${{H}_{2}}S{{O}_{4}}$ required to make 0.1N (${{N}_{2}}$) of solution in 1 litre (${{V}_{2}}$) be ‘${{V}_{1}}$’.
Using normality equation, we have
  & {{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}} \\
 & {{V}_{1}}=\dfrac{{{N}_{2}}{{V}_{2}}}{{{N}_{1}}} \\
 & {{V}_{1}}=\dfrac{0.1N\times 1000mL}{36.05N}=2.77mL \\

 Therefore, the volume of 96% solution of ${{H}_{2}}S{{O}_{4}}$(36.05 N) required to make 0.1N solution in 1 litre (1000 mL) is 2.77 mL.

(c) Using normality equation, we can calculate the volume (${{V}_{2}}$) to which 10 mL (${{V}_{1}}$) 96% solution of ${{H}_{2}}S{{O}_{4}}$(${{N}_{1}}=36.05N$) is diluted to prepare solution of 2 N (${{N}_{2}}$).
  & {{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}} \\
 & {{V}_{2}}=\dfrac{{{N}_{1}}{{V}_{1}}}{{{N}_{2}}} \\
 & {{V}_{1}}=\dfrac{36.05N\times 10mL}{2N}=180.25mL \\

Therefore, 10 mL of 36.05N should be diluted to 180.25 mL to make 2N solution.

Note: Convert all the units for volume in either litre or millilitre and use the same unit throughout the calculation to avoid errors. Carefully follow the steps to calculate the volume of the solution from the specific gravity given.