The volume of 1.00 kg of the solution is 885.76 cm3.
Calculation
The molar volume of the mixture:
V_mix = x_B * V_B + (1 - x_B) * V_A
Where:
V_mix is the molar volume of the mixture (cm3/mol)
x_B is the mole fraction of CHCl3 (0.4693)
V_B is the partial molar volume of CHCl3 (80.235 cm3/mol)
V_A is the partial molar volume of acetone (74.166 cm3/mol)
Calculate the mass of one mole of the mixture:
M_mix = (x_B * 119.59 + (1 - x_B) * 58.08) # g/mol
Where:
M_mix is the mass of one mole of the mixture (g/mol)
Calculate the number of moles in 1.00 kg of the solution:
n_mix = 1000 / M_mix # mol
Where:
n_mix is the number of moles in 1.00 kg of the solution (mol)
Calculate the volume of 1.00 kg of the solution:
V_solution = n_mix * V_mix # cm3
Where:
V_solution is the volume of 1.00 kg of the solution (cm3) .
Question
The partial molar volumes of acetone (propanone) and chloroform (trichloromethane) in a mixture in which the mole fraction of chcl 3 is 0.4693 are 74.166 cm 3 /mol and 80.235 cm 3 /mol, respectively. what is the volume of a solution of mass 1.000 kg ?