Question

The density of a solution containing 13% by mass of sulphuric acid is 1.09 g/ml. calculate the molarity and the normality of the solution.

Answer 1

Hint: Molarity and normality are units of concentration in chemistry.
Molarity is defined as the number of moles of a solute dissolved in 1L of a solution.
Normality: Normality is defined as the number of equivalents per litre of solution. In other words, it is a measure of concentration that is equal to gram equivalent weight of solute per litre of solution.

Complete step by step answer:
Given, Density of the solution is = 1.09 g/ml
Mass of sulphuric Acid = 13%
$\because $ let the mass of the solution is 100gm.
So, we can say that mass of sulphuric Acid = 13gm
We know, Density \[ = {\text{ }}mass/volume\]

Volume = $\dfrac{{Mass}}{{Density}}$

Volume = $\dfrac{{100}}{{1.09 \times 1000}}$ $\left[ {\because 1L = 1000ml} \right]$

Or $Volume = \dfrac{1}{{10.9}}L$
Molecular weight of sulphuric acid = 98g/mol
Moles of sulphuric acid =$\dfrac{{13}}{{98}} moles$

$\therefore Molarity = \dfrac{{Moles{\text{ }}of{\text{ }}solute{\text{ }}}}{{Volume{\text{ }}of{\text{ }}solution}} = \dfrac{{13}}{{98}} \times \dfrac{{10.9}}{1} = 1.445M$

To find normality we have to know the equivalent weight of ${H_2}S{O_4}$
${H_2}S{O_4} \to 2{H^ + } + S{P_4}^{2 - }$
Basicity \[ = {\text{ }}2\]
So the equivalent weight is = $\dfrac{{Molecular{\text{ }}weight}}{2} = \dfrac{{98}}{2} = 49$
$\therefore Normality = \dfrac{{\% strength{\text{ }}of{\text{ }}solution{\text{ }} \times {\text{ }}density{\text{ }}of{\text{ }}solute}}{{Equivalent{\text{ }}Weight}} = \dfrac{{13 \times 10.9}}{{49 \times 1}} = 2.89N$

Hence, the molarity of a solution is 1.445M and the normality of a solution is 2.89N.

Note: Relation between molarity and normality is $N = M \times n$, where ‘n’ is the no. of equivalents
For ${H_2}S{O_4}1M = 2N$
From this relation, we can easily get the normality from molarity of a solution.