Answer:
5.3%
Explanation:
Let the volume be 1 L
volume , V = 1 L
use:
number of mol,
n = Molarity * Volume
= 0.8846*1
= 0.8846 mol
Molar mass of CH3COOH,
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 CH3COOH,
m = number of mol * molar mass
= 0.8846 mol * 60.05 g/mol
= 53.12 g
volume of solution = 1 L = 1000 mL
density of solution = 1.00 g/mL
Use:
mass of solution = density * volume
= 1.00 g/mL * 1000 mL
= 1000 g
Now use:
mass % of acetic acid = mass of acetic acid * 100 / mass of solution
= 53.12 * 100 / 1000
= 5.312 %
≅ 5.3%
The mass percentage of acetic acid is 5.3%. It can be calculated by subsituting the values in Molarity formula.
It is defined as the ratio of the Number of moles of solute over Volume of solution in litres.
n = Molarity * Volume
n= 0.8846*1
n= 0.8846 mol
Calculation for Mass of CH₃COOH:
Molar mass of CH₃COOH = 60.052 g/mol
m = number of mol * molar mass
m= 0.8846 mol * 60.05 g/mol
m= 53.12 g
Volume of solution = 1 L = 1000 mL
Density of solution = 1.00 g/mL
Mass of solution = density * volume
= 1.00 g/mL * 1000 mL
Mass of solution= 1000 g
Mass % of acetic acid = mass of acetic acid * 100 / mass of solution
= 53.12 * 100 / 1000
Mass % of acetic acid= 5.312 %
Find more information about Molarity here:
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