Commercially available, concentrated hydrochloric acid contains 38% HC

1.	Commercially available, concentrated hydrochloric acid contains 38% HCl by mass, (a) What is the molarity of this solution? The density is 1.19 g mL–1 (b) What volume of concentrated HCl is required to make 1.00 L of 0.10 M HCI?
Answer» (a) We know that, Molarity = \(\frac{Number\,of\,moles\,of\,solute(HCl)}{volume\,of\,solution}\) Number of moles of HCl = \(\frac{Mass\,of\,the\,solute(HCl)}{Molecular\,mass\,of\,HCl}\) = \(\frac{38}{36.5}\) Volume of solution: 38 g of HCl is present in 100 g of the solution Volume of 100 g of the acid = \(\frac{Mass}{Density}\) = \(\frac{100\,g}{1.19\,g\,cm^{-3}}\) [\(\because\) 1mL = 1cm³] = 84.033 ≈ 84.05 cm³ Molarity = \(\frac{\frac{38}{36.5}}{\frac{84.05}{1000}}\)^{= 12.38 M} (b) For monobasic acid M₁V₁ = M₂V₂ 12.38 M × V₁ = 10 M × 1000 mL V₁ = \(\frac{10\,M\times1000\,mL}{12.38\,M}\) = 807.754 ≈ 807.8 mL

Commercially available, concentrated hydrochloric acid contains 38% HCl by mass, (a) What is the molarity of this solution? The density is 1.19 g mL–1 (b) What volume of concentrated HCl is required to make 1.00 L of 0.10 M HCI?

Answer»

(a) We know that,

Molarity = \(\frac{Number\,of\,moles\,of\,solute(HCl)}{volume\,of\,solution}\)

Number of moles of HCl

= \(\frac{Mass\,of\,the\,solute(HCl)}{Molecular\,mass\,of\,HCl}\) = \(\frac{38}{36.5}\)

Volume of solution:

38 g of HCl is present in 100 g of the solution

Volume of 100 g of the acid = \(\frac{Mass}{Density}\) = \(\frac{100\,g}{1.19\,g\,cm^{-3}}\) [\(\because\) 1mL = 1cm³]

= 84.033 ≈ 84.05 cm³

Molarity = \(\frac{\frac{38}{36.5}}{\frac{84.05}{1000}}\)^{= 12.38 M}

(b) For monobasic acid

M₁V₁ = M₂V₂

12.38 M × V₁ = 10 M × 1000 mL

V₁ = \(\frac{10\,M\times1000\,mL}{12.38\,M}\)

= 807.754 ≈ 807.8 mL