Question

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

Answer 1

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

Complete step by step answer:
Given that the w/w measure of ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ stock solution is 61.25%. That means, 61.25 g is of ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ 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 $1.6\,{\text{g}}\,{\text{m}}{{\text{L}}^{ - 1}}$.
So,
${\text{Density}} = \dfrac{{{\text{Mass}}}}
{{{\text{Volume}}}}$
$ \Rightarrow 1.6\,{\text{g}}\,{\text{m}}{{\text{L}}^{ - 1}} = \dfrac{{100\,{\text{g}}}}
{{{\text{Volume}}}}$
$ \Rightarrow {\text{Volume}} = {\text{62}}{\text{.5}}\,{\text{mL}}$
So, the volume of stock solution is $62.5\,{\text{mL}}$.
Now, we have to calculate the moles of solute $\left( {{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}} \right)$. The mass of solute is 61.25 g and molar mass of solute is $98\,{\text{g}}\,{\text{mo}}{{\text{l}}^{ - 1}}$.
Moles of solute=$\dfrac{{{\text{Mass}}}}
{{{\text{Molar}}\,{\text{mass}}}}$
\[ \Rightarrow {\text{Moles}}\,{\text{of}}\,{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} = \dfrac{{61.25\,{\text{g}}}}
{{98\,{\text{g}}\,{\text{mo}}{{\text{l}}^{ - 1}}}}\]
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.
${M_1} = \dfrac{{61.25}}
{{98}} \times \dfrac{{1000}}
{{62.5}}$
\[\]
${M_2} = 0.3\,{\text{M}}$=
${V_2} = 200\,{\text{mL}} = \dfrac{{200}}
{{1000}}{\text{L}}$
$\dfrac{{61.25}}
{{98}} \times \dfrac{{1000}}
{{62.5}} \times {V_1} = 0.3\, \times \dfrac{{200}}
{{1000}}$
$ \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 ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$.

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.