Question

The density \[({\text{in g m}}{{\text{l}}^{ - 1}})\] of a \[3.60{\text{M}}\] sulphuric acid solution having \[29\% {\text{ }}{{\text{H}}_2}{\text{S}}{{\text{O}}_4}\] \[[{\text{molar mass}} = 98{\text{ g mo}}{{\text{l}}^{ - 1}}]\] by mass will be:
A.\[1.64\]
B.\[1.88\]
C.\[1.22\]
D.\[1.45\]

Answer 1

Hint: Using the formula of mass percentage we will get mass of sulphuric acid and number of moles can be calculated. Using the formula for molarity of the solution we will calculate the density of solution.
Formula used: \[{\text{Molarity}} = \dfrac{{{\text{no}}{\text{. of moles }}}}{{{\text{volume of solution}}({\text{ml}})}} \times 1000\]
\[{\text{no}}{\text{. of moles}} = \dfrac{{{\text{mass}}}}{{{\text{molar mass}}}}\]

Complete step by step answer:
Sometimes the concentration is defined in terms of percentage. There are two ways of representing the concentration in terms of percentage that is percentage by mass and percentage by volume. When either of them is not specified in the question we will consider that percentage by mass is given to us.
Percentage by mass is defined as the mass of solute that is present in 100 gram of solution.
Mass percentage is of \[{{\text{H}}_2}{\text{S}}{{\text{O}}_4}\] is given to us that is \[29\% \] , so 29 g of \[{{\text{H}}_2}{\text{S}}{{\text{O}}_4}\] is contained in 100 g of the solution.
\[{\text{moles of }}{{\text{H}}_2}{\text{S}}{{\text{O}}_4} = \dfrac{{{\text{mass}}}}{{{\text{molar mass}}}}\]
Putting the values we will get the number of moles of sulphuric acid as:
\[{\text{moles of }}{{\text{H}}_2}{\text{S}}{{\text{O}}_4} = \dfrac{{29}}{{{\text{98}}}} = 0.2959{\text{ mole}}\]
Mass of \[{{\text{H}}_2}{\text{S}}{{\text{O}}_4}\] is 29 g. Molar mass of \[{{\text{H}}_2}{\text{S}}{{\text{O}}_4}\] is \[2 \times 1 + 32 + 4 \times 16 = 98\].
We know\[{\text{density}} = \dfrac{{{\text{mass}}}}{{{\text{volume}}}}\].
Rearranging we will get:
\[{\text{volume of solution}} = \dfrac{{{\text{mass of solution}}}}{{{\text{density of solution}}}}\].
Substituting the values in molarity formula we will get,
\[{\text{Molarity}} = \dfrac{{{\text{moles of }}{{\text{H}}_2}{\text{S}}{{\text{O}}_4}}}{{{\text{mass of solution}}}} \times {\text{density of solution}} \times 1000\]
Again we can rearrange the equation to get more simplified version as:
\[{\text{density of solution}} = \dfrac{{{\text{Molarity}} \times {\text{mass of solution}}}}{{{\text{moles of }}{{\text{H}}_2}{\text{S}}{{\text{O}}_4} \times 1000}}\]
\[ \Rightarrow {\text{density}} = \dfrac{{3.6 \times 100}}{{0.2959 \times 1000}} = 1.22{\text{g m}}{{\text{l}}^{ - 1}}\]

Hence, the correct option is C.

Note:
Sulphuric acid is an inorganic acid or mineral acid. It contains Sulphur, hydrogen and oxygen. It is a colourless and odorless liquid and is very corrosive in nature. It is soluble in water. Another name for sulphuric acid is Vitriol. It is synthesized by various processes such as contact process, lead chamber process and wet sulphuric acid process.