# (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?

Answer

Verified

335.1k+ views

**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

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

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

**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

\[\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.

**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.

Last updated date: 27th Sep 2023

•

Total views: 335.1k

•

Views today: 4.35k

Recently Updated Pages

What do you mean by public facilities

Difference between hardware and software

Disadvantages of Advertising

10 Advantages and Disadvantages of Plastic

What do you mean by Endemic Species

What is the Botanical Name of Dog , Cat , Turmeric , Mushroom , Palm

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

What is the IUPAC name of CH3CH CH COOH A 2Butenoic class 11 chemistry CBSE

Drive an expression for the electric field due to an class 12 physics CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference Between Plant Cell and Animal Cell

The dimensions of potential gradient are A MLT 3A 1 class 11 physics CBSE

Define electric potential and write down its dimen class 9 physics CBSE

Why is the electric field perpendicular to the equipotential class 12 physics CBSE