Question

Calculate the molarity of \[20\% \] \[\dfrac{w}{w}\] aqueous solution of sulphuric acid.

Answer 1

Hint: Sulphuric acid is an ionic compound containing two hydrogen ions and one sulphate ion in a single molecule. Its weight by weight \[\dfrac{w}{w}\] percentage can be used to calculate the molarity by calculating the number of moles using its molecular weight.

Complete answer:
Concentration on any solution is a way of measuring the amount of solute that is present in a fix amount of solvent. There are different methods and formulas of calculating and expressing concentrations of the solutions. One such method is a weight by weight \[\dfrac{w}{w}\] percentage that utilizes the mass of both solute and solvent in grams.
The weight percentage is determined by finding out the ratio of the mass of solute that is dissolved and the total mass of the solution obtained upon the dissolution of solute in solvent. The formula can be written as follows:
\[\dfrac{w}{w} = \dfrac{{{\text{mass of solute}}}}{{{\text{mass of solution}}}} \times 100\]
A weight percentage of \[20\% \] suggests that there is \[20g\] solute dissolved in \[100g\] of the solution. Therefore there is \[20g\] of sulphuric acid present in \[100g\] of solution.
The molar mass of sulphuric acid is \[98gmo{l^{ - 1}}\] which can be used to calculate its amount in moles. The density of water and its mass can be used to calculate the volume of water present.
\[{\text{moles of }}{H_2}S{O_4} = \dfrac{{{\text{given mass}}}}{{{\text{molar mass}}}} = \dfrac{{20g}}{{98gmo{l^{ - 1}}}} \approx 0.20moles\]
The density of water is \[1gm{L^{ - 1}}\] and the ratio of mass and density of water gives its final volume:
\[volume = \dfrac{{{\text{mass of water}}}}{{density}} = \dfrac{{100g}}{{1gm{L^{ - 1}}}} = 100ml = 0.1L\]
Molarity is another form of expressing concentration that is calculated by dividing the number of moles of solute by the volume of solution in liters.
\[ \Rightarrow \] Thus, the molarity of sulphuric acid is :
\[molarity = \dfrac{{{\text{moles of solute}}}}{{{\text{volume of solution(in L)}}}} = \dfrac{{0.02mol}}{{0.1L}} = 0.002M\]

Note:
In the given question we assume that the density of solution is the same as that of the pure water without any solvent in it. The volume of the solvent is not the same as the volume of the solution but in some cases they are assumed to be the same to simplify calculations.