Question

Describe how you prepare $1$ L of $0.50$ M solution of sulphuric acid from $5.0$M solution of sulphuric acid?

Answer 1

To answer this question we should know about dilution. How we do dilution, the effect of dilution. Dilution means increases the concentration of solvent in the solution which is generally water. We will determine the volume of the sulphuric acid solution which is required for dilution.

Complete solution:
$5.0$M solution of sulphuric acid means $5.0$ mol of sulphuric acid is dissolved in one litre water.
$0.50$M solution of sulphuric acid means $0.50$mol of sulphuric acid is dissolved in one litre water.
To prepare $1$ L of $0.50$M solution from $5.0$M solution, we have to take some volume of $5.0$M solution then we have to dilute it or we can say we have to add water to make it up to $1$ L so, the molarity of the new solution will be $0.50$M.
The formula to calculate the concentration or volume on dilution is as follows:
${{\text{M}}_{\text{1}}}{{\text{V}}_{\text{1}}}\,{\text{ = }}\,{{\text{M}}_{\text{2}}}{{\text{V}}_{\text{2}}}$
Where,
${{\text{M}}_{\text{1}}}$ is the molarity of the solution before dilution
${{\text{V}}_{\text{1}}}$ is the volume of the solution before dilution
${{\text{M}}_{\text{2}}}$ is the molarity of the solution after dilution
${{\text{V}}_2}$ is the volume of the solution after dilution
On substituting $5.0$ M for ${{\text{M}}_{\text{1}}}$, $0.50$ M for ${{\text{M}}_{\text{2}}}$ and $1$ L for ${{\text{V}}_2}$,
${\text{5}}{\text{.0}}\, \times \,{{\text{V}}_{\text{1}}}\,{\text{ = }}\,0.50\,\, \times \,\,1{\text{L}}$
$\,{{\text{V}}_{\text{1}}}\,{\text{ = }}\,\dfrac{{0.50\,\, \times \,\,1{\text{L}}}}{{5.0}}$
$\,{{\text{V}}_{\text{1}}}\,{\text{ = }}\,0.1\,{\text{L}}$
We need $0.1$L of $5.0$ M sulphuric acid solution.
Therefore, to prepare $1$ L of $0.50$ M solution of sulphuric acid from $5.0$M solution of sulphuric acid we will take $0.1$ of $5.0$ M sulphuric acid solution and make it up to one litre.

Note: When we add water in a solution, the concentration of the solute per unit volume decreases. On dilution, concentration decreases. The addition of water is also known as dilution. We calculated the volume in litre. It can also be calculated in mL. One liter is equal to a thousand millilitres. So, practically we will take hundred millilitre for dilution.