Question

Conc. ${{H}_{2}}S{{O}_{4}}$ has a density of 1.98g/ml and is 98% ${{H}_{2}}S{{O}_{4}}$ by weight. Its normality is?
(a) 2N
(b) 19.8N
(c) 39.6N
(d) 98N

Answer 1

Hint Normality of a solution is defined as the number of equivalent of solute present in one litre of (1000 ml) solution and is given by $N=\dfrac{W\times 1000}{E\times V(ml)}$ and molarity is the number of moles of solute in one litre of solution and is given by, $M=\dfrac{n}{V(Litre)}$.

Complete Step by step solution:
In the lower classes, we have studied the calculation of molarity, molality, normality and many others. We shall see these in detail.
Normality of a solution is defined as the number of equivalent of solute present in one litre of (1000 ml) solution and is given by $N=\dfrac{W\times 1000}{E\times V(ml)}$
Suppose a solution is prepared by the dissolving W gm of solute of equivalent weight ‘E’ in ‘V’ ml.
Number of gram equivalent of solute =$\dfrac{W}{E}$ and ‘V’ ml of solution have $\dfrac{W}{E}$ equivalent of solute, so 100 ml of solution have,
Normality $N=\dfrac{W\times 1000}{E\times V(ml)}$
\[\Rightarrow Normality=Molarity\times f\] ……………………..(i)
where $f$ is the valency factor.
Molarity is defined as the number of moles of solute present in the one litter of solution. If a solution have $n$ number of mole of solute particle dissolved in ‘V’ ml of solution so its molarity will be –
Molarity $M=\dfrac{n}{V(Litre)}=\dfrac{n\times 1000}{VmL}$ ………………..(ii)
98% ${{H}_{2}}S{{O}_{4}}$ by weight mean 100gm of ${{H}_{2}}S{{O}_{4}}$ solution contains 98gm of ${{H}_{2}}S{{O}_{4}}$. To calculate normality of the acid solution first we will calculate molarity of the solution.
To calculate molarity, it requires a number of moles and volume of the solution.
Mole of ${{H}_{2}}S{{O}_{4}}$ is $mole(n)=\dfrac{mass}{molar-mass}$,
By substituting mass = 98gm and molar mass of ${{H}_{2}}S{{O}_{4}}$= 98gm in this formula we get,
\[n=\dfrac{98}{98}=1\]
So, volume of 100gm of ${{H}_{2}}S{{O}_{4}}$ is,
\[Density(d)=\dfrac{mass(m)}{Volume(V)}\]
By rearranging and substituting the values of the data we have,
\[V=\dfrac{100g}{1.98gm{{l}^{-1}}}=50.5ml\]
It means 98gm ${{H}_{2}}S{{O}_{4}}$ is present in the 50.5ml of the solution. So molarity of the solution from equation …. (ii) that is $M=\dfrac{n}{V(Litre)}=\dfrac{n\times 1000}{VmL}$
So, after putting n=1 and V=50.5ml in this equation, we get
\[M=\dfrac{1}{50.5ml}\times 1000=19.8mol/L\]
So, to calculate normality we will put the value of molarity and factor in the equation… (i)
Since, the basicity of ${{H}_{2}}S{{O}_{4}}$ is two. So, its factor will be two.
Now, by putting f=2 and M=19.8 in this equation we will get normality of the solution and it will be,
\[Normality=19.8\times 2=39.8\]

So, option (c) will be the correct option.

Note: Note that the factors in case of acid and bases are respectively the basicity and acidity. In oxidant and reductant it is the change in oxidation number par mole of an atom in a redox reaction.