Hydrochloric acid is usually purchased in a concentrated form that is 37.0% HCl by mass and has a density of 1.20 g/mL. How much concentrated solution would it take to prepare 3.00 L of 0.480 M HCl upon dilution with water?

Sagot :

The required volume of concentrated stock solution of hydrochloric acid is 102.71 mL.

What is the relation between the volume & density?

Volume of any solution will be calculated by using the density as:

Volume = Mass/Density

Given that percent mass of HCl is 37%, so mass of the HCl = 37g

So that mass of solution = 100g

And density of solution = 1.2 g/mL

Volume of solution = 100/1.2 = 83.33 mL = 0.08333 L

Moles of HCl will be calculated as:

n = W/M, where

W = given mass = 37g

M = molar mass = 36.4g/mol

n = 37/36.4 = 1.014mol

Then the molarity of the stock solution will be calculated as:

M = n/V

M = 1.014 / 0.0833 = 12.17M

Now the required volume of stock solution will be calculated by using the following equation as:

M₁V₁ = M₂V₂, where

M₁ & V₁ is the molarity and volume of stock solution and M₂ & V₂ is the molarity and volume of dilute solution. On putting values in the above solution we get

12.17 × V₁ = 0.5 × 2500

12.17 × V₁ = 1250

V₁ = 102.71 mL

Hence required volume of stock solution is 102.71 mL.

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