The molar concentration of acetic acid in this vinegar is ?
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Answer:
The molecular weight of acetic acid is 60.05 grams/mol. So 12.4g/60.05g/mol = 0.206 moles. And there's 0.248 liters in 248 mL to to ger mols per liter divide 0.206 moles by 0.248 liters and: 0.833 mol/L.
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Other answers
An analysis of vinegar showed that 12.4 g of acetic acid, CH3COOH(l), were present in 248 mL of vinegar sample. The molar concentration of acetic acid in this vinegar is Molar concentration = moles of solute ÷ liters of solution Mass of 1 mole of CH3COOH = 12 + 3 + 12 + 32 + 1 = 60 Moles of CH3COOH = 12.4 ÷ 60 Liters of solution = 248/1000 = 0.248 Molar concentration = (12.4 ÷ 60) ÷ 0.248 Molar concentration = 0.8333 mole per liter
An analysis of vinegar showed that 12.4 g of acetic acid, CH3COOH(l), were present in 248 mL of vinegar sample. The molar concentration of acetic acid in this vinegar is Molar concentration = moles of solute ÷ liters of solution Mass of 1 mole of CH3COOH = 12 + 3 + 12 + 32 + 1 = 60 Moles of CH3COOH = 12.4 ÷ 60 Liters of solution = 248/1000 = 0.248 Molar concentration = (12.4 ÷ 60) ÷ 0.248 Molar concentration = 0.8333 mole per liter
electron1
The molecular weight of acetic acid is 60.05 grams/mol. So 12.4g/60.05g/mol = 0.206 moles. And there's 0.248 liters in 248 mL to to ger mols per liter divide 0.206 moles by 0.248 liters and: 0.833 mol/L.
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