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Mass of NaoH in 43 g ml of 0.1098 M NOOH = moles times molecular weight times volume of NaoH in lines = 0.1098 times 40 times 40.90/1000 = .0179 g Moles of base of equivalence point = mass/mole. Ut = 0.179/40 = 0.00449 M Moles at out of equivalent point CH_3COOH + NaoH rightarrow CH_3 COONa + H_2 O therefore moles of acid of equivalence point = 0.00449 M mass of acid = moles times molecular weight = 0.00449 times 60.05 = 0.269 g volume of concentrated vinegar = mass/density = 0.269/1.008 = 0.267 mL mass of vinegar solution = 25 g (since mass of vinegar is negligible density of loiter is 1 so 25 ml solution = 25 g) Mass % of acetic acid = .0269/25 times 100 = 1.076 %
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