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AnnZ [28]
3 years ago
9

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?

Chemistry
1 answer:
jok3333 [9.3K]3 years ago
4 0

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.

Molality = \frac{moles}{mass of solvent}

Substitute the values into above formula as follows.

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

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

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THE EMPIRICAL FORMULA OF THE SUBSTANCE IS C2H5NO

Explanation:

The steps involved in calculating the empirical formula of this substance in shown in the table below:

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1. % Composition              40.66              8.53                 23.72                27.09

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3. Divide by smallest

value (0.6931)          3.3883/1.6931    8.53/1.6931    1.6943/1.6931   1.6931/1.6931

                               =      2.001                  5.038           1.0007                      1

4. Whole number ratio        2                       5                   1                               1

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