Question

The species mass densities of a three?component (A, B, and C) liquid mixture are: acetone, $\rho$ A= 326.4, kg/m³, acetic acid, $\rho$ B = 326.4 kg/m³, and ethanol, $\rho$ C = 217.6 kg/m³.

Determine the following for this mixture:
1. The mass fraction of each species in the mixture.

2. The mole fraction of each species in the mixture.

3. The mass of each component required to make one cubic meter of mixture.

Answer 1

Given:

The species mass densities of a three‐component (A, B, and C) liquid mixture are: acetone, $\rho$ _A= 326.4, kg/m³, acetic acid, $\rho$ _B = 326.4 kg/m³, and ethanol, $\rho$ _C = 217.6 kg/m^3.

Assume 1 m³ of total mixture

1. mass fraction of each mixture:

Mass of acetone in the mixture W_A = 326.4 kg

Mass of acetic acid in the mixtureW_B= 326.4 kg

Mass of ethanol in the mixtureW_C= 217.6 kg

then, Total mass of the mixture will be W_T = 870.4 kg

Mass frsction of acetone m_A = W_A/W_T = 326.4/870.4 = 0.375

Mass fraction of acetic acid m_B = W_B/W_T= 326.4/870.4= 0.375

Mass fraction of ethanol m_C= W_C/W_T= 217.6/870.4= 0.25

2. Mole fraction of each component:

Molar mass of acetone M_A = 58.08 g/mol

Molar mass of acetic acid M_B = 60.05 g/mol

Molar mass of ethanol M_C= 46.068 g/mol

Calculate of no. moles of each component:

Moles of acetone n_A = WA/MA = 326.4/58.08 = 5.62 mol

Moles of acetic acid nB = 326.4/60.05= 5.44 mol

Moles of ethanol nC= WC/MC = 217.6/46.068= 4.72 mol

Total no of moles nT = 15.78 mol

mole fraction of acetone xA = nA/nT = 5.62/15.78= 0.356

mole fraction of acetic acid x_B= n_B/nT= 5.44/15.78= 0.345

mole fraction of ethanol x_C= n_C/n_T= 4.72/15.778= 0.299

3. Mass required of each component to make 1 m³ of mixture will be=density of component * 1 m³

   Hence mass of each component required will be 326.4 kg of acetone, 326.4 kg of acetic acid & 217.6 kg of ethanol.