Commercially available concentrated Hydrochloric acid contains $38\,\% \,HCl$ by mass. What is the molarity of the solution, density of the solution is $1.19\,g/ml$. What is the volume of $HCl$ is required to make $10\,L$ of $0.10\,M\,HCl$.

Hint:We know that,
The mass of solute in one liter of solution is defined as molarity. It is the ideal concentration unit for stoichiometry calculations. The formula is,
$Molarity = \dfrac{{Mass\,of\,solute(in\,moles)}}{{Volume\,of\,solution\,(in\,litres)}}$

Complete step by step answer:

First, calculate the volume of the solution from density.
Given: The density of the solution is $1.19\,g/ml$.
We know that the density is defined as the mass per unit volume and it is used to measure the rigidity of the material.
$Density{\text{ }} = \dfrac{{Mass}}{{Volume}}$
Substituting the value of density of solution we get,
$Volume = \dfrac{{100}}{{1.19}} = 84\,ml$
Now, calculate the number of moles of hydrochloric acid in the solution.
Given: The mass of the solution is $38\,g$.
We know the molecular mass of hydrochloric acid is $36.5\,g/mol.$
We know that the amount of moles in a given amount of any substance is equal to the grams of the substance divided by its molecular weight.
The mathematically expressed as,
$Mole = \dfrac{{weight\,of\,the\,subs\tan ce}}{{\,Molecular\,weight}}$
The number of moles of hydrochloric acid is calculated as,
$Moles = \dfrac{{38\,g}}{{36.5\,g/mol}} = 1.04\,moles$
Now, calculate the molarity of the solution,
$Molarity = \dfrac{{1.04\,moles}}{{0.084\,L}} = 12.38\,M$
The molarity of the solution is $12.38\,M$.
The volume of $HCl$ is required to make $10\,L$ of an $0.10\,M\,HCl$ can be calculated as,
$Mass = \,number\,of\,moles \times molecular\,weight$
Substituting the number of moles and molecular weight we get,
$Mass = \,0.1\,mol \times 36.5\,g/mol = 3.65\,g$
Let us assume that the mass of $HCl$ in $3.65\,g$ as Y.
$\dfrac{{38}}{{100}} = \dfrac{{3.65}}{Y}$
$Y = 9.6\,g$
The volume of $HCl$ is required $ = \dfrac{{9.6}}{{1.19}} = 8.067\,ml$
The volume of $HCl$ is required to make $10\,L$ of an $0.10\,M\,HCl$ is $8.067\,ml.$

Note:
Molarity is defined as the number of moles of solute in one liter of solution and molality is used to measure the moles to the kilogram of the solvent.
The mathematical expression of molality is,
\[Molality{\text{ }}\left( m \right){\text{ }} = {\text{ }}\dfrac{{moles{\text{ }}of{\text{ }}solute\left( {Mol} \right)}}{{kilograms{\text{ }}of{\text{ }}solvent\left( {Kg} \right)}}{\text{ }}\].