Hydrochloric acid is usually purchased in concentrated form with a 37.0% HCl concentration by mass and a density of 1.20 g / mL. How much of the concentrated stock solution in milliliters should you use to make 2.5 L of 0.500 M HCl

The volume of the concentrated stock solution needed to prepare the solution is 102.71 mL

Let the mass of the concentrated HCl solution be 100 g.

thus, the mass of 37.0% HCl will be 37 g

Mass of solution = 100 g

Density of solution = 1.20 g/ mL

Mass of HCl = 37 g

Molar mass of HCl = 1 + 35.5 = 36.5 g/mol

Mole of HCl = 1.014 mole

Volume of solution = 83.33 mL = 83.33 / 1000 = 0.08333 L

Molarity of stock solution (M₁) = 12.17 M

Volume of diluted solution (V₂) = 2.5 L = 2.5 × 1000 = 2500 mL

Molarity of diluted solution (M₂) = 0.5 M

M₁V₁ = M₂V₂

12.17 × V₁ = 0.5 × 2500

12.17 × V₁ = 1250

Divide both side by 12.17

V₁ = 1250 / 12.17

Therefore, the volume of the concentrated stock solution needed to prepare the solution is 102.71 mL

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0.10 L

Explanation:

Given: 37 gram of HCl are dissolved in 100 g of solution.

Density of solution= 1.20 g/ml

Volume of solution =

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

where,

n = moles of solute =

= volume of solution in ml = 83.3 ml

0.10 L of concentrated stock solution in milliliters should you use to make 2.5 L of 0.500 M HCl

2h2o2--> 2h2o + o2

this reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen.

add 2 molecules of hydrogen peroxide and 2 molecules of water. since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

this may not be entirely correct but let me know.

is there any options? maybe im just not seeing them but i need the options to

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