A domestic cleanser contains concentrated HCl as the active ingredient. A titration experiment was carried out to determine the concentration of HCl in the domestic cleanser. 20.0 ml of the cleanser were diluted to 250.0 ml with distilled water. 25.0 ml of the diluted cleanser were the titrated against 0.5 mol L-1 sodium hydroxide solution. 15.00 ml of sodium hydroxide solution were required for complete neutralization.

(A)Calculate the molarity of hydrochloric acid in the cleanser.
(B)Calculate the concentration, in g dm-3, of hydrochloric acid in the cleanser.

Thank you

so the Molarity of the diluted is M/(250/20)

25ml*(M/(250/20)=15ml*.5

M= 15.*.5*250/20*1/25=15*.25 check that.
from preceeding,
molarity= gramssolute/molmass*1liter
grams solute/liter= M*molmassHCl
= 15*.25/36

To calculate the molarity of hydrochloric acid (HCl) in the cleanser, we can use the principle of stoichiometry and the equation of the neutralization reaction between HCl and sodium hydroxide (NaOH).

Given data:
Volume of diluted cleanser used for titration (V₁) = 25.0 ml = 0.025 L
Volume of NaOH solution required for neutralization (V₂) = 15.00 ml = 0.015 L
Concentration of NaOH solution (C₂) = 0.5 mol L⁻¹

Since the balanced chemical equation for the reaction between HCl and NaOH is 1:1, we can equate the moles of HCl with the moles of NaOH involved in the reaction.

Moles of NaOH used = Concentration of NaOH × Volume of NaOH
Moles of NaOH used = 0.5 mol L⁻¹ × 0.015 L
Moles of NaOH used = 0.0075 mol

Since the balanced equation is 1:1 between HCl and NaOH, the moles of HCl present in the cleanser will also be 0.0075 mol.

Now we can calculate the volume of the diluted cleanser used for titration in terms of the original undiluted cleanser:

Volume of undiluted cleanser (V) = Volume of diluted cleanser × Dilution factor
Dilution factor = Total volume of diluted cleanser / Volume of undiluted cleanser
Dilution factor = 250.0 ml / 20.0 ml
Dilution factor = 12.5

Therefore, the volume of the undiluted cleanser used for titration = 0.025 L / 12.5
Volume of undiluted cleanser used for titration = 0.002 L

Molarity of HCl in the cleanser (C) = Moles of HCl / Volume of undiluted cleanser
Molarity of HCl in the cleanser = 0.0075 mol / 0.002 L
Molarity of HCl in the cleanser = 3.75 mol L⁻¹

Answer: The molarity of hydrochloric acid in the cleanser is 3.75 mol L⁻¹.

To calculate the concentration of hydrochloric acid (HCl) in the cleanser in g dm⁻³ (grams per decimeter cubed), we need to use the molarity of HCl and its molar mass.

The molar mass of HCl is 36.461 g mol⁻¹ (1 mol of HCl weighs approximately 36.461 grams).

Concentration of HCl in the cleanser (in g dm⁻³) = Molarity of HCl × Molar mass of HCl
Concentration of HCl in the cleanser = 3.75 mol L⁻¹ × 36.461 g mol⁻¹

Concentration of HCl in the cleanser = 136.72 g dm⁻³

Answer: The concentration of hydrochloric acid in the cleanser is 136.72 g dm⁻³.