Question

\[{\rm{135}}\,\,{\rm{mL}}\]of \[{\rm{0}}{\rm{.224}}\,\,{\rm{M}}\]\[{\rm{MgS}}{{\rm{O}}_4}\]solution was heated until the solution becomes \[{\rm{105}}\,\,{\rm{mL}}\]due to evaporation of water. Calculate the molarity of the concentrated solution.

Answer 1

Hint: Molarity (M) may be defined as the number of moles of solute dissolved per litre of the solution.
Mathematically, \[{\rm{molarity}}\,{\rm{(M)}}\,{\rm{ = }}\dfrac{{{\rm{number}}\,\,{\rm{of}}\,\,{\rm{moles}}\,\,{\rm{of}}\,\,{\rm{solute}}}}{{{\rm{volume}}\,\,{\rm{of}}\,\,{\rm{solution}}\,\,{\rm{in}}\,\,{\rm{litres}}}}\]
The units of molarity are moles per litre \[{\rm{(mol}}{{\rm{L}}^{{\rm{ - 1}}}}{\rm{)}}\]or moles per cubic decimetre \[{\rm{(mold}}{{\rm{m}}^{{\rm{ - 3}}}}{\rm{)}}\]. The symbol used to designate molar concentration is \[{\rm{M}}\].

Formula used \[{{\rm{M}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}{\rm{(before}}\,\,{\rm{dilution)}} = {{\rm{M}}_{\rm{2}}}{{\rm{V}}_{\rm{2}}}{\rm{(after}}\,\,{\rm{dilution)}}\]
where, \[{{\rm{M}}_{\rm{1}}}\]and \[{{\rm{M}}_{\rm{2}}}\]are the molarities (concentrations) of the solution before and after dilution respectively.
\[{{\rm{V}}_{\rm{1}}}\]and \[{{\rm{V}}_{\rm{2}}}\]are the volumes of the solution before and after dilution respectively.

Complete Step by Step Solution:
It is important to remember that when a solution is diluted with a solvent, the number of moles of solute in solution remain unchanged.

So, number of moles of solute = molarity \[ \times \]volume of solution in litres
Thus, for two different concentrations:
\[{{\rm{M}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}{\rm{(before}}\,\,{\rm{dilution)}} = {{\rm{M}}_{\rm{2}}}{{\rm{V}}_2}{\rm{(after}}\,\,{\rm{dilution)}}\]

It is called the molarity equation. In case of balanced equation if \[{{\rm{n}}_1}\]moles of one reactant combines with \[{{\rm{n}}_2}\]moles of the second reactant, then
\[\dfrac{{{{\rm{M}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}}}{{{{\rm{n}}_1}}} = \dfrac{{{{\rm{M}}_{\rm{2}}}{{\rm{V}}_2}}}{{{{\rm{n}}_2}}}\]

As per given data,
Molarity of \[{\rm{MgS}}{{\rm{O}}_4}\]solution, \[{{\rm{M}}_{\rm{1}}}\]\[{\rm{ = }}\,\,{\rm{0}}{\rm{.224}}\,\,{\rm{M}}\]
Volume of \[{\rm{MgS}}{{\rm{O}}_4}\]solution, \[{{\rm{V}}_{\rm{1}}}\]\[{\rm{ = }}\,\,{\rm{135}}\,\,{\rm{mL}}\]

Molarity of concentrated solution, \[{{\rm{M}}_{\rm{2}}}\]\[ = \,\,?\]
Volume of concentrated solution, \[{{\rm{V}}_2}\] \[{\rm{ = }}\,\,{\rm{105}}\,\,{\rm{mL}}\]
Use the molarity equation to solve for the value of \[{{\rm{M}}_{\rm{2}}}\]as shown below:
\[\begin{array}{l}{{\rm{M}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}{\rm{(before}}\,\,{\rm{dilution)}} = {{\rm{M}}_{\rm{2}}}{{\rm{V}}_2}{\rm{(after}}\,\,{\rm{dilution)}}\\ \Rightarrow {{\rm{M}}_{\rm{2}}} = \dfrac{{{{\rm{M}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}}}{{{{\rm{V}}_2}}}\\ \Rightarrow {{\rm{M}}_{\rm{2}}} = \dfrac{{{\rm{0}}{\rm{.224}}\,\,{\rm{M}}\,\, \times \,\,{\rm{135}}\,\,{\rm{mL}}}}{{{\rm{105}}\,\,{\rm{mL}}}}\\ \Rightarrow {{\rm{M}}_{\rm{2}}} = \dfrac{{30.24}}{{105}}{\rm{M}}\\ \Rightarrow {{\rm{M}}_{\rm{2}}} = \,\,0.288\,\,{\rm{M}}\end{array}\]

Hence, molarity of the concentrated solution is found to be \[0.288\,\,{\rm{M}}\]
Therefore, the answer is \[0.288\,\,{\rm{M}}\]

Note: Most of the reactions carried out in the laboratory take place quantitatively in solutions. Thus, it is important to know the amount of the reacting substances present in solution. Knowing the concentration of the solution in terms of normality, molarity etc., the amount of substance in a given volume can be calculated. The only disadvantage of this concentration unit i.e., molarity is that its value changes with the change in temperature.