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Question: 61.25% (w/w) \({{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}\) stock solution has a densi...

61.25% (w/w) H2SO4{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} stock solution has a density of 1.6 g mL−11.6\,{\text{g}}\,{\text{m}}{{\text{L}}^{ - 1}}. The molecular weight of H2SO4{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} is 98 g mol−198\,{\text{g}}\,{\text{mo}}{{\text{l}}^{ - 1}}. The volume (mL) of stock solution required to prepare a 200 mL solution of 0.3 M H2SO4{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} is

Explanation

Solution

Here, first we have to calculate the volume of H2SO4{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} stock solution using the formula, Density=MassVolume{\text{Density}} = \dfrac{{{\text{Mass}}}} {{{\text{Volume}}}}. Then, we have to calculate the molarity of stock solution. Then we have to use the formula M1V1=M2V2{M_1}{V_1} = {M_2}{V_2}to calculate the volume of stock solution required to prepare a 200 mL solution of 0.3 M H2SO4{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}.

Complete step by step answer:
Given that the w/w measure of H2SO4{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} stock solution is 61.25%. That means, 61.25 g is of H2SO4{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} is present in 100 g of solution. So,
Mass of solute=61.25 g
Mass of solution=100 g
Now, we have to calculate the volume of solution.
The density of the stock solution is given as 1.6 g mL−11.6\,{\text{g}}\,{\text{m}}{{\text{L}}^{ - 1}}.
So,
Density=MassVolume{\text{Density}} = \dfrac{{{\text{Mass}}}} {{{\text{Volume}}}}
⇒1.6 g mL−1=100 gVolume\Rightarrow 1.6\,{\text{g}}\,{\text{m}}{{\text{L}}^{ - 1}} = \dfrac{{100\,{\text{g}}}} {{{\text{Volume}}}}
⇒Volume=62.5 mL\Rightarrow {\text{Volume}} = {\text{62}}{\text{.5}}\,{\text{mL}}
So, the volume of stock solution is 62.5 mL62.5\,{\text{mL}}.
Now, we have to calculate the moles of solute (H2SO4)\left( {{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}} \right). The mass of solute is 61.25 g and molar mass of solute is 98 g mol−198\,{\text{g}}\,{\text{mo}}{{\text{l}}^{ - 1}}.
Moles of solute=MassMolar mass\dfrac{{{\text{Mass}}}} {{{\text{Molar}}\,{\text{mass}}}}

{{98\,{\text{g}}\,{\text{mo}}{{\text{l}}^{ - 1}}}}$$ Now, we have to calculate the molarity of stock solution. Molarity=$\dfrac{{{\text{Moles}}\,{\text{of}}\,{\text{solute}}}} {{{\text{Volume}}\,{\text{of}}\,{\text{solution}}}}$ $ \Rightarrow {\text{Molarity}} = \dfrac{{\dfrac{{61.25}} {{98}}}} {{\dfrac{{62.5}} {{1000}}}}$ Now, we have to calculate the volume of stock solution of $\left( {{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}} \right)$needed to to prepare a 200 mL solution of 0.3 M ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$. So, we have to use the formula ${M_1}{V_1} = {M_2}{V_2}$, where ${M_1}$ is the molarity of original stock solution and ${M_2}$ is the molarity of newly formed stock solution. ${V_1}$ is the volume of the original stock solution and ${V_2}$is the volume of the newly formed stock solution. ${M_1} = \dfrac{{61.25}} {{98}} \times \dfrac{{1000}} {{62.5}}$

M2=0.3 M{M_2} = 0.3\,{\text{M}}=
V2=200 mL=2001000L{V_2} = 200\,{\text{mL}} = \dfrac{{200}} {{1000}}{\text{L}}
61.2598×100062.5×V1=0.3 ×2001000\dfrac{{61.25}} {{98}} \times \dfrac{{1000}} {{62.5}} \times {V_1} = 0.3\, \times \dfrac{{200}} {{1000}}
⇒V1=0.3×200×98×62.51000×61.25×1000=3675612500=0.006 L=6 mL\Rightarrow {V_1} = \dfrac{{0.3 \times 200 \times 98 \times 62.5}} {{1000 \times 61.25 \times 1000}} = \dfrac{{3675}} {{612500}} = 0.006\,{\text{L}} = {\text{6}}\,{\text{mL}}
Therefore, 6 mL of stock solution is required to prepare a 200 mL solution of 0.3 M H2SO4{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}.

Note: It is to be noted that w/w% is a way of expressing concentration of a solution. For example, 2% w/w solution HCl means that 2 g of HCl is present in 100 g of solution.