Question

\[98\% {\text{ }}{H_2}S{O_4}\] solution by mass and has a density of \[1.80g.m{L^{ - 1}}\]. Volume of acid require to make \[1{\text{ }}litre\]of \[0.1M{\text{ }}{H_2}S{O_4}\]solution is:

Answer 1

Hint: We can define molarity as the number of moles of solute per litre of solution. Also \[98\% {\text{ }}{H_2}S{O_4}\] means \[98g\] of acid in \[100g\] of solution.

Complete step by step answer: Let’s start by understanding the concept of molarity; molarity is defined as the number of moles of solute per litre of solution. So, if we are talking about \[1M\] solution of \[HCl\], then it means that 1 mole of \[HCl\] is dissolved in 1 litre of solution, which would mean that \[36.5g\] of \[HCl\] is being dissolved in 1litre of solution. Solution will be made up of both the solvent and the solute.
Now, coming back to the question, we are given \[98\% {\text{ }}{H_2}S{O_4}\] which would mean \[98g{\text{ }}of{\text{ }}{H_2}S{O_4}\] per \[100g\] of \[{H_2}S{O_4}\] solution. We are also given the density of the \[{H_2}S{O_4}\] which is \[1.80g.m{L^{ - 1}}\].
will have \[1{\text{ }}mole{\text{ }}of{\text{ }}{H_2}S{O_4}\] so, the mass of \[{H_2}S{O_4}\] will be \[98g\].
So, for \[0.1M{\text{ }}{H_2}S{O_4}\] solution, the mass of \[{H_2}S{O_4}\] will be \[9.8g\]
So, the amount to be added in \[mL\] will be \[\dfrac{{Mass{\text{ }}to{\text{ }}be{\text{ }}added}}{{Density}} = \dfrac{{9.8}}{{1.80}} = 5.55mL\]
So, we need to add in \[994.45mL\] water to make \[1{\text{ }}litre{\text{ }}of{\text{ }}0.1M{\text{ }}{H_2}S{O_4}\].

Note: We can use the concept of molarity in chemical and related industry for a long time. Wherever there is any need of chemicals or there is laboratory the concept of Molarity is being used. It has been used for standardizing the solutions so that they can be used for laboratory as well as manufacturing purposes.