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Tema [17]
4 years ago
9

If 11.5 ml of vinegar sample (d=1g/ml) is titrated with 18.5 ml of standardized Sodium hydroxide

Chemistry
1 answer:
Oksi-84 [34.3K]4 years ago
4 0

Answer:

10.35 M

Explanation:

Molarity = moles solute/ L solution

18.5 mL = 0.0185 L solution

11.5 mL (1g/1mL)= 11.5 grams vinegar (1 mol / 60.052 grams) = 0.1915 mol vinegar

M = 0.1915 mol vinegar/ 0.0185 L solution = 10.35M

* Please text me at 561-400-5105 for tutoring: I can do assignments, labs, exams, etc. :)

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