The mass percent of the methanol is 21.83%.
The parameter
- Methanol volume = 200.0 ml
- The density of methanol = 0.7920 g/ml
- Water volume = 567.1 ml
- The density of water = 1.000 g/ml
Calculate the mass of methanol and water.
[tex]\rho = \frac{m}{V}[/tex]
m = ρ × V
Where m is a mass (kg), ρ is the density (kg/m³), and V is volume (m³)
- m₁ = mass of methanol = 0.7920×200.0
m₁ = 158.4 grams - m₂ = mass of water = 1.000×567.1
m₂ = 567.1 grams
Mass percent of the methanol
= [tex]\frac{mass \: of \: methanol}{mass \: of \: solution} \times 100\%[/tex]
= [tex]\frac{m_1}{m_1 + m_2} \times 100\%[/tex]
= [tex]\frac{158.4}{158.4 + 567.1} \times 100\%[/tex]
= [tex]\frac{158.4}{725.5} \times 100\%[/tex]
= 0.2183 × 100%
= 21.83%
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