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

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"