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500 mL of 0.90 M acetic acid (CH3COOH) is combined with 600 mL of 12% w/v acetic acid. Calculate the resulting molarity of acetic acid in the mixed solution. (Molar mass of CH3COOH = 60 g/mol)

1. 0.90 M
2. 1.20 M
3. 1.50 M
4. 1.80 M

Correct answer: 1.50 M

Solution

Moles = 0.45 + 1.20 = 1.65 in total volume 1.1 L, giving 1.5 M.

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