Question

How do you prepare a $1\mathrm{M}$ $\mathrm{HCl}$ solution, which has a specific gravity of $1.18$? The HCl solution has $35\%$ purity. Calculate the volume required to prepare the solution.

Answer 1

Step 1: Mass of the solution to make the $35\%$ pure

Let us assume that we have to prepare $1\mathrm{L}$of $1\mathrm{mol}{\mathrm{L}}^{-1}$ $\mathrm{HCl}$.

$1\mathrm{M}$ represents the molarity (number of moles of solute per litre of solution) of the $\mathrm{HCl}$ solution.

That is, $1\mathrm{M}$ $\mathrm{HCl}$ is one mole $\mathrm{HCl}$ in $1\mathrm{L}$ of water

The mass of Hydrogen is $1\mathrm{g}$.

The mass of Chlorine is $35.5\mathrm{g}$.

Therefore, the molar mass of $\mathrm{HCl}$ is $35.5+1=36.5g$

$35\%$ purity of the $\mathrm{HCl}$ solution means $35g$ of $\mathrm{HCl}$ is present

Thus, the mass of the solution to make $35\%$ purity will be $36.5\mathrm{g}\times \frac{100\mathrm{g}}{35\mathrm{g}}=104.2\mathrm{g}$

Step 2: The volume required

The Specific Gravity of a substance is the ratio of its mass to that of an equal volume of water at the same temperature and pressure.

That is, $\mathrm{Specific}\mathrm{gravity}=\frac{\mathrm{mass}\mathrm{of}\mathrm{a}\mathrm{volume}\text{'}\mathrm{V}\text{'}\mathrm{of}\mathrm{a}\mathrm{material}}{\mathrm{amount}\mathrm{of}\mathrm{a}\mathrm{volume}\text{'}\mathrm{V}\text{'}}$

By rearranging this equation, the amount of volume required to make the $1\mathrm{M}$ $\mathrm{HCl}$ of solution have the specific gravity $1.18$ is;

$$\begin{gathered} \mathrm{Amount}\mathrm{of}\mathrm{a}\mathrm{volume}\text{'}\mathrm{V}\text{'}=\frac{\mathrm{mass}\mathrm{of}\mathrm{a}\mathrm{volume}\text{'}\mathrm{V}\text{'}\mathrm{of}\mathrm{a}\mathrm{material}}{s\mathrm{pecific}\mathrm{gravity}} \\ \mathrm{Amount}\mathrm{of}\mathrm{a}\mathrm{volume}\text{'}\mathrm{V}\text{'}=\frac{104.2}{1.18} \\ \mathrm{Amount}\mathrm{of}\mathrm{a}\mathrm{volume}\text{'}\mathrm{V}\text{'}=88\mathrm{mL} \end{gathered}$$.

$\frac{104.2}{1.18}=88\mathrm{mL}$

Therefore, the volume required to prepare $1\mathrm{M}$ $\mathrm{HCl}$ solution, which has a specific gravity of $1.18$ is $88\mathrm{mL}$