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Question

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Calculate the mass of acetic acid (CH₃COOH)) needed for orepairing a 250ml solution with a molar concentration of 0,25mole/L. Estimate the percentage of the total mass of acetic acid in the prepared solution (p=1,12g/ml)

Answer 1

250 ml is the same as 0.250 L. This can be seen as 250 ml x 1 L / 1000 ml = 0.250 L (dimensional analysis)

To find the mass of CH₃COOH, we need to use dimensional analysis again:

0.250 L x 0.25 mol / L = 0.0625 mols CH3COOH

Molar mass CH₃COOH = 60.1 g/mol

0.0625 mols x 60.1 g/mol = 3.76 g CH₃COOH = mass CH₃COOH

p=1.12g/ml means the density of the solution is 1.12 g/ml

Since we have 250 ml of solution, we can find the total mass of this solution using this density:

250 ml x 1.12 g / ml = 280 g

The percentage of the total mass that is CH3COOH is 3.76 g / 280 g (x100%) = 1.34%

Answer 2

You need 250ml solution with 0.25 moles / L

1 l contains 0.25 moles so 250 ml contains (250/1000) x 0.25 moles

The mass of the prepared solution = pV = 1.12 x 250 g

The molecular mass of acetic acid = 2 x atomic mass of C + 4 x atomic mass of H + 2 x atomic mass of O.

You can look these up and calculate this. Let's call this molecular mass M.

1 mole of acetic acid has this mass M (in g)

The mass you need (lets call it x) will be (250/1000) x 0.25 x M

Mass percentage = (mass of acid) x 100 / (mass of solution) = 100x / (1.12 x 250)