Question

How many mL of sulphuric acid of density $1.84\,{\text{g}}\,{\text{m}}{{\text{L}}^{ - 1}}$ containing 95.6 mass % of ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$
should be added to one litre of 40 mass% solution of ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ of density 1.31 ${\text{g}}\,{\text{m}}{{\text{L}}^{ - 1}}$ in order to prepare 50 mass% solution of sulphuric acid of density $1.40\,{\text{g}}\,{\text{m}}{{\text{L}}^{ - 1}}$?
A.66.2 mL
B.55.2 mL
C.95.5 mL
D.105.2 mL

Answer 1

Here, first we have to calculate the volume of 95.6 mass % of ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ solution. Then, we have to calculate the mass solute needed to make 40 mass % of ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ solution. Then, we have to calculate the solute needed to make a 50 mass % solution.

Complete step by step answer:
First we have to calculate the volume of 95.6 mass % of ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$solution using the formula of density. 95.6 mass % of ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ means the mass of ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ is 95.6 g and mass of the solution is 100 g. The density given is $1.84\,{\text{g}}\,{\text{m}}{{\text{L}}^{ - 1}}$.
${\text{Density = }}\dfrac{{{\text{Mass}}}} {{{\text{Volume}}}}$
$\Rightarrow 1.84\,{\text{g}}\,{\text{m}}{{\text{L}}^{ - 1}} = \dfrac{{100\,{\text{g}}}} {{{\text{Volume}}}}$

{{1.84}} = 54.35\,{\text{mL}}$$ Now, for the 40% mass solution of ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$, Volume=1 L=1000 ml Density=1.31 ${\text{g}}\,{\text{m}}{{\text{L}}^{ - 1}}$ So, mass of the solution=$ = 1.31\,{\text{g}}\,{\text{m}}{{\text{L}}^{ - 1}} \times 1000\,{\text{mL = 1310}}\,{\text{mL}}$ The mass percent is given as 40%. Let’s take the mass of the solute to be x. Mass percent=$\dfrac{{{\text{Mass}}\,{\text{of}}\,{\text{solute}}}} {{{\text{Mass}}\,{\text{of}}\,{\text{solution}}}} \times 100$ $ \Rightarrow 40 = \dfrac{x} {{1310\,{\text{mL}}}} \times 100$ $ \Rightarrow x = 40 \times \dfrac{{1310}} {{100}} = 524\,{\text{g}}$ Now, to make mass percent 50 solution of density 1.40 g/mL , Mass of solution=1 L=1000 mL Density=1.40 g/mL Volume=1400 mL Solute present$ = \dfrac{{50 \times 1400}} {{100}} = 700\,{\text{g}}$ So, the mass of solute needs to be added to make 50% from 40% by mass=700-524=176 g. Now, we have to calculate the volume of ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ needs to be added to 40%. Volume=$\dfrac{{{\text{Mass}}}} {{{\text{Density}}}} = \dfrac{{176\,}} {{1.84\,}} = 95.6521\,{\text{mL}}$ So, volume of ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ required to be added is 95.6521 ml. **So, the correct answer is Option C.** **Note:** It is to be noted that mass percent of a solute is defined as the number of parts by mass of one component (solute or solvent) per 100 parts by mass of the solution. Concentration can also be expressed in terms of weight/volume. For example, 10% solution of KCl (W/V) means that 10 g of the salt is dissolved in 100 mL of the solution.