What is the molality of an acetic acid solution that contains 0.500 mol of HC2H3O2 (molar mass: 60g/mol) in 0.800 kg of water?

Respuesta :

Answer: The molality of given acetic acid solution is 0.625 m.

Explanation:

Given: Moles of solute = 0.5 mol

Mass of solvent = 0.8 kg

Molality is the number of moles of solute present in a kg of solvent.

Therefore, molality of the given solution is calculated as follows.

[tex]Molality = \frac{moles}{mass of solvent}[/tex]

Substitute the values into above formula as follows.

[tex]Molality = \frac{moles}{mass of solvent}\\= \frac{0.5 mol}{0.8 kg}\\= 0.625 m[/tex]

Thus, we can conclude that the molality of given acetic acid solution is 0.625 m.

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