Question

$29.2$% (w/w) ${\text{HCl}}$stock solution has a density of ${\text{1}}{\text{.25}}\,{\text{g}}\,{\text{m}}{{\text{l}}^{ - 1}}$. The molecular weight of ${\text{HCl}}$is ${\text{36}}{\text{.5}}\,{\text{g}}\,{\text{mo}}{{\text{l}}^{ - 1}}$. The volume (ml) of stock solution required to prepare a $200$ml solution $0.4\,{\text{M}}$ ${\text{HCl}}$is:
A. ${\text{5}}\,{\text{mL}}$
B. ${\text{6}}\,{\text{mL}}$
C. $8\,{\text{mL}}$
D. $15\,{\text{mL}}$

Answer 1

We will determine the number of mole of solute (hydrochloric acid) by using the mole formula. Then the amount of solvent (water) in kg and then molarity of the solution. Then by comparing the molarity and volume product we can determine the volume of the stock solution.

Complete Step by step answer: $29.2$% (w/w) ${\text{HCl}}$ stock means that $29.2$ gram of hydrochloric acid is present in $100$ gram of solution.
Determine the number of mole of hydrochloric acid as follows:
${\text{Mole}}\,{\text{ = }}\,\dfrac{{{\text{Mass}}}}{{{\text{Molar}}\,{\text{mass}}}}$
Substitute ${\text{36}}{\text{.5}}\,{\text{g}}\,{\text{mo}}{{\text{l}}^{ - 1}}$ for molar mass and $29.2$ gram for mass.
${\text{Mole}}\,{\text{ = }}\,\dfrac{{29.2\,\,{\text{g}}}}{{{\text{36}}{\text{.5}}\,{\text{g}}\,{\text{mo}}{{\text{l}}^{ - 1}}}}$
${\text{Mole}}\,{\text{ = }}\,0.8\,{\text{mol}}$
So, the mole of hydrochloric acid is$0.8$.
Use the density formula to determine the volume of the solution as follows
${\text{density}}\,{\text{ = }}\,\dfrac{{{\text{mass}}}}{{{\text{volume}}}}$
Substitute ${\text{1}}{\text{.25}}\,{\text{g}}\,{\text{m}}{{\text{l}}^{ - 1}}$for density and $100$ gram for mass of the solution.
${\text{1}}{\text{.25}}\,{\text{g}}\,{\text{m}}{{\text{l}}^{ - 1}}\,{\text{ = }}\,\dfrac{{{\text{100}}\,{\text{g}}}}{{{\text{volume}}}}$
${\text{volume = }}\,\dfrac{{{\text{100}}\,{\text{g}}}}{{{\text{1}}{\text{.25}}\,{\text{g}}\,{\text{m}}{{\text{l}}^{ - 1}}}}$
$\Rightarrow {\text{volume = }}\,80\,{\text{ml}}$
Convert the volume of solution from ml to L as follows:
$1000\,{\text{ml}}\,{\text{ = }}\,{\text{1}}\,{\text{L}}$
$\Rightarrow 80\,{\text{ml}}\,{\text{ = }}\,0.08\,{\text{L}}$
The formula of molarity is as follows:
${\text{Molarity}}\,{\text{ = }}\,\dfrac{{{\text{Moles}}\,{\text{of}}\,{\text{solute}}}}{{{\text{L of solution}}}}$
Substitute $0.8$ for moles of solute and $0.08$ ml for volume of solution.
${\text{Molarity}}\,{\text{ = }}\,\dfrac{{0.8\,{\text{mol}}}}{{0.08\,{\text{L}}}}$
$\Rightarrow {\text{Molarity}}\,{\text{ = }}\,{\text{10}}\,{\text{M}}$
So, the molarity of the stock solution is $10\,{\text{M}}$.
Now we will determine the volume of $10\,{\text{M}}$ ${\text{HCl}}$ stock solution required to prepare the $200$ml of$0.4\,{\text{M}}$ ${\text{HCl}}$ as follows:
${{\text{M}}_{\text{1}}}{{\text{V}}_{\text{1}}}\,{\text{ = }}\,{{\text{M}}_{\text{2}}}{{\text{V}}_{\text{2}}}$
Where,
${{\text{M}}_{\text{1}}}$ is the molarity of the solution having
${{\text{V}}_{\text{1}}}$ volume.
${{\text{M}}_{\text{2}}}$ is the molarity of the solution having ${{\text{V}}_{\text{2}}}$ volume.
Substitute $10\,{\text{M}}$for${{\text{M}}_{\text{1}}}$, $200$ml for ${{\text{V}}_{\text{2}}}$and $0.4\,{\text{M}}$ for ${{\text{M}}_{\text{2}}}$.
$\Rightarrow 10\,{\text{M}}\,\, \times \,\,{{\text{V}}_1}\,{\text{ = }}\,0.4\,{\text{M}}\,\, \times \,\,{\text{200}}\,{\text{ml}}$
$\Rightarrow {{\text{V}}_1}\,{\text{ = }}\dfrac{{\,0.4\,{\text{M}}\,\, \times \,\,{\text{200}}\,{\text{ml}}}}{{10\,{\text{M}}\,}}$
$\Rightarrow {{\text{V}}_1}\,{\text{ = }}8\,{\text{ml}}$
So, $8$ml of $10\,{\text{M}}$ stock solution is required to prepare a $200$ml solution$0.4\,{\text{M}}$ ${\text{HCl}}$.

Therefore, option (C) $8\,{\text{mL}}$is correct.

Note: If the molar mass of the solute is not given, it can be calculated by adding the mass of each atom of the compound. The unit of molarity is mol/L so, it is necessary to convert the unit of solution volume from ml to L. The unit of molarity is represented by M. The small ‘m’ represents the unit of molality.