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Xelga [282]
3 years ago
15

The distinctive odor of vinegar is due to acetic acid, HC₂H₃O₂, which reacts with sodium hydroxide. If a 3.45 mL sample of vineg

ar needs 42.5 mL of 0.115 M NaOH to be neutralized, how many grams of acetic acid are in a 1.00 L bottle of this vinegar?​
Chemistry
1 answer:
Vilka [71]3 years ago
8 0

Answer:

Explanation:

CH₃COOH + NaOH = CH₃COONa + H₂O .

42.5 mL of .115 M of NaOH will contain .0425 x .115 moles of NaOH

= 48.875 x 10⁻⁴ moles NaOH

It will react with same number of moles of acetic acid

So number of moles of acetic acid in 3.45 mL = 48.875 x 10⁻⁴

number of moles of acetic acid in 1000 mL = 48.875 x 10⁻⁴ x 10³ / 3.45 moles

= 1.4167 moles

= 1.4167 x 60 gram

= 85 grams .

So 85 grams of acetic acid will be contained in one litre of acetic acid.

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2 years ago
Calculate the freezing point (in degrees C) of a solution made by dissolving 7.99 g of anthracene{C14H10} in 79 g of benzene. Th
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<u>Answer:</u> The freezing point of solution is 2.6°C

<u>Explanation:</u>

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

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m_{solute} = Given mass of solute (anthracene) = 7.99 g

M_{solute} = Molar mass of solute (anthracene) = 178.23  g/mol

W_{solvent} = Mass of solvent (benzene) = 79 g

Putting values in above equation, we get:

5.5-\text{Freezing point of solution}=1\times 5.12^oC/m\times \frac{7.99\times 1000}{178.23g/mol\times 79}\\\\\text{Freezing point of solution}=2.6^oC

Hence, the freezing point of solution is 2.6°C

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