Comercially available concentrated HCL contains 38% HCL by mass.

i . What is molarity of the solution.(d=1.19ml).

ii . What volume of conc. HCL is required to make 1L of 0.1 M HCL.

38% by mass solution of HCl means that 38 g of HCl is present in 62g of water, thereby making the total mass of solution equal to 100g. The molar mass of HCl is 36.5g. Therefore, the number of moles of HCl that are present in 38g of HCl are

= mass of HCl / molar mass of HCl
= 38 / 36.5
= 1.04
We known, Molarity = Number of moles/ Volume of solution (in L)
As Volume of solutions is not known therefore, we will take help of density

Density = mass / volume
Mass of solution = Mass of solute + Mass of solvent
= 100 g
Volume = 100/ 1.19
= 84.033 mL

​On substituting the above values in the formula of molarity, we will get
M = 1.04 ×1000/ 84.033
= 12.37 M

​2.
The molarity of 38 % solution is 12.37 M
As we are supposed to make it 0.10 M in 1 L, therefore we can use the formula
M1V1 = M2V2 ...(ii)
Where 1 stands for commercially available HCl and 2 stands for final solution.

M1 = 12.37M
M2 = 0.10 M
V2 = 1 L
on substituing the above value in equation (ii), we will get
V1 = 0.10 × 1 / 12.37
= 0.00808 L
or 8.08 mL of HCl would be required.