Question

Density of a 2.05M solution of acetic acid in watee is 1.02g/ml .The molality of tge solution is

Answer 1

Assume there is 100mL of solution.

Find the number of moles of acetic acid using its molarity.

Find the mass of the solution using its density.

Find the mass of the acetic acid using (1) and molar mass of acetic acid.

Find the mass of solvent by minusing (2) by (3).

Find the molality by dividing (1) by (4).

The molality should be [math]2.29[/math] [math]mol[/math] [math]kg^{-1}[/math]

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Answer 2

Answer:

  m = 2.285mol kg⁻¹

Explanation:

molality - m

molarity - M

Density - d

M₂ - Molecular mass - CH₃COOH = 60g

m = 1000M / 1000d - MM₂

m = 1000 × 2.05 / (1000 × 1.02) - (2.05 × 60)

m = 2050 / 1020 - 123

m = 2050 / 897

m = 2.285mol kg⁻¹