Determine the mass percent composition of a solution made by
dissolving 4.91 moles CH3OH in 3.01 L of
H2O
Molar mass of CH3OH,
MM = 1*MM(C) + 4*MM(H) + 1*MM(O)
= 1*12.01 + 4*1.008 + 1*16.0
= 32.042 g/mol
use:
mass of CH3OH,
m = number of mol * molar mass
= 4.91 mol * 32.04 g/mol
= 157 g
Density of water is 1 Kg/L
So,
Mass of water = density * volume
= 1 Kg/L * 3.01 L
= 3.01 Kg
= 3010 g
Use:
Mass % of CH3OH = mass of CH3OH * 100 / mass of solution
= 157 * 100 / (157 + 3010)
= 4.96 %
Answer: 4.96 %
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