What is the weight percent (%) acetic acid (CH3CO2H) for a 5.88 m aqueous acetic acid solution?
Consider 1 Kg of solvent
use:
molality = number of mol of CH3CO2H / mass of solvent in Kg
5.88 = number of mol of CH3CO2H / 1 Kg
number of mol of CH3CO2H = 5.88 mol
Molar mass of CH3CO2H,
MM = 2*MM(C) + 4*MM(H) + 2*MM(O)
= 2*12.01 + 4*1.008 + 2*16.0
= 60.052 g/mol
use:
mass of CH3CO2H,
m = number of mol * molar mass
= 5.88 mol * 60.05 g/mol
= 353.1 g
mass % = mass of CH3CO2H * 100 / mass of solution
= 353.1 * 100 / (353.1 + 1000)
= 26.1 %
Answer: 26.1 %
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