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(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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\[\text{Normality}=\dfrac{\text{weight of solute (in grams)}}{\text{equivalent weight}}\times \dfrac{1000}{\text{volume of solution (in mL)}}\]

if normality of one solution is given, that of other can be calculated using normality equation which is given as

\[{{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}\]

Specific gravity is the ratio of the density of the given substance to the density of water. density of water is 1$gm{{L}^{-1}}$.

(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

\[\begin{align}

& \text{Equivalent weight = }\dfrac{\text{Molar mass}}{\text{Basicity}}=\dfrac{98g}{2}=49g\,e{{q}^{-1}} \\

& \\

\end{align}\]

Given specific gravity of ${{H}_{2}}S{{O}_{4}}$= 1.84. specific gravity is given as

\[\begin{align}

& \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}} \\

\end{align}\]

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

\[\begin{align}

& \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 \\

\end{align}\]

(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

\[\begin{align}

& {{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 \\

\end{align}\]

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 (${{N}_{2}}$).

\[\begin{align}

& {{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 \\

\end{align}\]

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