Question

Equal volumes of \[30\% \] by mass of \[{{\text{H}}_2}{\text{S}}{{\text{O}}_4}\] (density \[ = 1.218{\text{ g m}}{{\text{l}}^{ - 1}}\]) and \[70\% \]by mass of \[{{\text{H}}_2}{\text{S}}{{\text{O}}_4}\] (density \[ = 1.810{\text{ g m}}{{\text{l}}^{ - 1}}\]). If the density of the mixture is \[1.425{\text{ g m}}{{\text{l}}^{ - 1}}\]. Calculate the molarity and molality of a solution.

Answer 1

Hint: First of all we will calculate the mass of \[{{\text{H}}_2}{\text{S}}{{\text{O}}_4}\] and solution in each case considering the initial volume as V mL. Using the formula we can calculate the number of moles and volume and hence the molality of solution. Mass of solvent can be calculated by subtracting mass of \[{{\text{H}}_2}{\text{S}}{{\text{O}}_4}\] from total mass of solution. To calculate masses use the density values.

Formula used: \[{\text{molarity }} = \dfrac{{{\text{number of moles of solute}}}}{{{\text{volume of solution in mL}}}} \times 1000\]
 \[{\text{molality }} = \dfrac{{{\text{number of moles of solute}}}}{{{\text{mass of solvent in gram}}}} \times 1000\]
\[{\text{Density }} = \dfrac{{{\text{mass}}}}{{{\text{volume}}}}\]
\[{\text{Number of moles }} = \dfrac{{{\text{mass}}}}{{{\text{molar mass}}}}\]

Complete step-by-step answer:
Let the volume of each solution be V mL.
For mixtue I
Given that sulphuric acid is 30 percent by mass, this means that 30 gram of \[{{\text{H}}_2}{\text{S}}{{\text{O}}_4}\]is present in 100 g of water. Density of the solution is \[1.218{\text{ g m}}{{\text{l}}^{ - 1}}\]. So mass of solution will be with volume V will be:
\[{\text{mass of solution }}1{\text{ }} = 1.218{\text{ V gram}}\]
100 gram of solution contain 30 gram of \[{{\text{H}}_2}{\text{S}}{{\text{O}}_4}\]
1 gram of solution contain \[\dfrac{{30}}{{100}}\]gram of \[{{\text{H}}_2}{\text{S}}{{\text{O}}_4}\]
\[1.218{\text{ V}}\] gram of solution contain \[\dfrac{{30}}{{100}} \times 1.218{\text{ V }} = 0.3659{\text{ V}}\]gram of \[{{\text{H}}_2}{\text{S}}{{\text{O}}_4}\]
For mixture II
Given that sulphuric acid is 70 percent by mass, this means that 70 gram of \[{{\text{H}}_2}{\text{S}}{{\text{O}}_4}\]is present in 100 g of water. Density of the solution is \[1.810{\text{ g m}}{{\text{l}}^{ - 1}}\]. So mass of solution will be with volume V will be:
\[{\text{mass of solution 2 }} = 1.810{\text{ V gram}}\]
100 gram of solution contain 70 gram of \[{{\text{H}}_2}{\text{S}}{{\text{O}}_4}\]
1 gram of solution contain \[\dfrac{{70}}{{100}}\]gram of \[{{\text{H}}_2}{\text{S}}{{\text{O}}_4}\]
\[1.810{\text{V}}\] gram of solution contain \[\dfrac{{70}}{{100}} \times 1.810{\text{ V }} = 1.267{\text{V}}\]gram of \[{{\text{H}}_2}{\text{S}}{{\text{O}}_4}\]
Total mass of \[{{\text{H}}_2}{\text{S}}{{\text{O}}_4}\] will be: \[0.3659{\text{V}} + 1.267{\text{V}} = 1.6329{\text{V}}\]
Molar mass of \[{{\text{H}}_2}{\text{S}}{{\text{O}}_4}\] is \[98{\text{ g}}/{\text{mol}}\]
Number of moles of \[{{\text{H}}_2}{\text{S}}{{\text{O}}_4}\]\[ = \dfrac{{1.6329{\text{V}}}}{{98}} = 0.01666{\text{ V}}\]
Total volume of solution after mixing will be: 2V
Using the formula for molarity:
\[{\text{Molarity}} = \dfrac{{0.01666{\text{V}}}}{{2{\text{V}}}} \times 1000 = 8.33{\text{M}}\]
Using the mass of solution calculated in the above two cases we can calculate the mass of mixture as:
Total mass of mixture is \[1.425{\text{ gm}}{{\text{L}}^{ - 1}} \times 2{\text{V}} = 2.85{\text{V g}}\]
Total mass of \[{{\text{H}}_2}{\text{S}}{{\text{O}}_4}\] is \[1.6329{\text{V}}\]
Mass of solvent that is water will be \[{\text{2}}{\text{.85V}} - 1.6329{\text{V}} = 1.2171{\text{V}}\]
Hence molality can be calculated as:
\[{\text{molality }} = \dfrac{{0.01666{\text{V}}}}{{{\text{1}}{\text{.2171V}}}} \times 1000 = 13.69{\text{ m}}\]
Hence Molarity of the solution is \[8.33{\text{ M}}\] and molality of the solution is \[13.69{\text{ m}}\].

Note:Percentage composition is defined in two ways percentage by mass and percentage by volume. In percentage by volume mass in gram is replaced by volume in mL, i.e. it is equal to mass of solute divided by volume of solution multiplied by 100.