The volume of 50% methanol solution (p = 0.64 g / ml) required for the interaction with 17.8 g
August 29, 2021 | education|
The volume of 50% methanol solution (p = 0.64 g / ml) required for the interaction with 17.8 g of aminopropionic acid.
Aminopropionic acid: CH3-CH (NH2) -COOH. Molecular formula: C3H7NO2. M = 89.09 g / mol.
Aminopropionic acid reacts with methanol to form an ester.
CH3-CH (NH2) -COOH + CH3OH = CH3-CH (NH2) -COOCH3 + H2O
n (amine acid) = m / M = 17.8 / 89.09 = 0.2 mol.
n (CH3OH) = n (amino acid) = 0.2 mol.
m (CH3OH) = n (CH3OH) * M = 0.2 * 32.04 = 6.4 g.
To get a 50% solution, you need to dilute the original solution 2 times.
m (50% p. CH3OH) = 6.4 * 2 = 12.8 g.
V (50% p. CH3OH) = m / p = 12.8 / 0.64 = 20 ml.
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