(4) If a 1.0 L sample of 5% vinegar was diluted with distilled water so that...

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(4) If a 1.0 L sample of 5% vinegar was diluted with distilled water so that the final volume was 2.4 L what would be the new

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Answer 1

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4. Concentration of vinegar = 5% = 5 g / 100 mL

In 1 L, we have : 5 / 100 * 1000 = 50 g vinegar

Moles of Vinegar = Mass / molar mass = 50 / (60) = 0.833

Using the relation : M * V = constant. By assuming final molarity = M, we get:

0.833 * 1 = M * 2.4

Solving, we get: M = 0.833 / 2.4 = 0.347 M which is the concentration in mol / dm³

5. This depends on the volume of NaOH needed to react with the acetic acid. Just use the relation : molarity * Volume of NaOH = Moles of Acetic acid

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