Question

What volume of \[96\% {H_2}S{O_4}\]​ solution (density \[1.83g/mL\] ) is required to prepare \[4\] litre of \[3.0M\] \[{H_2}S{O_4}\] solution?

Answer 1

Hint: The percentage of a solution indicates the weight of solute present in \[100\;g\] of the solution. The volume of \[96\% \]\[{H_2}S{O_4}\] solution required to prepare \[4L\] of \[3M\]\[{H_2}S{O_4}\] solution can be calculated by determining the molarity of \[96\% {H_2}S{O_4}\] solution.

Complete step by step answer:
First of all, \[96\% {H_2}S{O_4}\] solution suggests that \[96g\] of \[{H_2}S{O_4}\] is present in \[100g\] of its solution.
The density of the solution is given as \[1.83g/mL\] .

The formula for density is given by the equation, \[density = \dfrac{{mass}}{{volume}}\].

Therefore, \[volume\;of\;the\;solution = \dfrac{{mass}}{{density}}\]

Mass of the solution \[ = 100g\]

Putting the values of mass and density in the formula, we get the volume of the solution as:

\[Volume = \dfrac{{100g}}{{1.83g/mL}}\]

The molarity of a solution is calculated by the formula:

\[Molarity = \dfrac{{Number\;of\;moles}}{{Volume\;of\;the\;solution\;in\;litres}}\]


Now, the number of moles is given as \[\dfrac{{Given\;weight}}{{Molecular\;weight}}\].

Given weight of \[{H_2}S{O_4} = 96g\]

Molecular weight of \[{H_2}S{O_4} = 2 \times Atomic\;weight\;of\;H + Atomic\;weight\;of\;S + 4 \times Atomic\;weight\;of\;O\]

\[ \Rightarrow Molecular\;weight\;of\;{H_2}S{O_4} = 2 \times 1\;g + 32\;g + 4 \times 16\;g = 2g + 32\;g + 64\;g = 98\;g\]

Therefore, number of moles of \[{H_2}S{O_4} = \dfrac{{96}}{{98}}\]

To convert the volume of the solution from \[mL\]to \[L\] (litres), we shall divide it by \[1000\] .

Therefore, volume of the solution in litres \[ = \dfrac{{100}}{{1.83 \times 1000}}L\]

Now, molarity \[ = \dfrac{{\dfrac{{96}}{{98}}}}{{\dfrac{{100}}{{1.83 \times 1000}}}}\]

\[ \Rightarrow molarity = \dfrac{{96 \times 1000 \times 1.83}}{{98 \times 100}}\]

\[ \Rightarrow molarity = 17.92M\]

The number of moles of \[{H_2}S{O_4}\] in both the solutions will remain the same.

From the formula of molarity, \[Number\;of\;moles = Molarity \times Volume\]

${M_1} = 17.92M$

${V_1} = ?$

${M_2} = 3M$

${V_2} = 4L$

Therefore, using \[{M_1}{V_1} = {M_2}{V_2}\] , we get,

\[17.92M \times {V_1} = 3M \times 4L\]

\[ \Rightarrow 17.92M \times {V_1} = 3M \times 4L\]

\[ \Rightarrow {V_1} = \dfrac{{12}}{{17.92}} = 0.6696L\]

\[ \Rightarrow {V_1} = 0.6696 \times 1000mL\]

\[ \Rightarrow {V_1} = 669.6mL = 670mL\]

Therefore, the volume of \[96\% {H_2}S{O_4}\] required to prepare a solution of \[4\] litre and \[3.0M\]is\[670mL\].

Note: When the percentage of a solution is mentioned, it implies the weight percentage of the solute in solution, and not the volume percentage.
Students must not forget to change the volume of solution in litres while calculating the molarity, to avoid making errors in calculation.
While calculating the density of the solution, students must be careful that they are required to consider the mass of the entire \[{H_2}S{O_4}\] solution \[(100g)\] , and not just the mass of the solute, \[{H_2}S{O_4}(96g)\].