A student needs to prepare 250.0 mL of a 0.175 mol/L solution of acetic acid. The starting material are pure acetic acid, CH3COOH, also known as acetic acid, glacial, a liquid with a density of 1.05 g/cm3, DI water and the required glassware.
What is the volume (in mL) of acetic acid, glacial, that she needs to measure into the 250-mL flask? (3 s.f.)
chem lab question!?
2018-03-13 1:21 pm
回答 (1)
2018-03-13 2:13 pm
No. of moles of CH₃COOH required = (0.175 mol/L) × (250.0/1000) = 0.04375 mol
Molar mass of CH₃COOH = (12.0×2 + 1.0×4 + 16.0×2) g/mol = 60.0 g/mol
Mass of CH₃COOH required = (0.04375 mol) × (60.0 g/mol) = 2.625 g
Density of CH₃COOH = 1.05 g/cm³
Volume of CH₃COOH required = (2.625 g) / (1.05 g/mol) = 2.50 cm³
Molar mass of CH₃COOH = (12.0×2 + 1.0×4 + 16.0×2) g/mol = 60.0 g/mol
Mass of CH₃COOH required = (0.04375 mol) × (60.0 g/mol) = 2.625 g
Density of CH₃COOH = 1.05 g/cm³
Volume of CH₃COOH required = (2.625 g) / (1.05 g/mol) = 2.50 cm³
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