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A 10 m3 mixture solution contains molecules A, B and C in water at ambient temperature.
Calculate:
(a) Each species mass concentration.
(b) Their molar concentration.
(c) Total mass concentration and molar concentration.
(d) Mass and mole fraction.
given:
Vmix= 10 m3
mA= 0.3 kg , mB= 0.5 kg , mC= 0.4 kg , mH2O= 9.8 kg
MWA= 50 , MWB= 250 , MWC= 10 , MWH2O= 18
Volume of total mixture = 10m3
Mass of A =0.3 kg
Mass of B = 0.5 kg
Mass of C = 0.4 kg
Mass of water = 9.8 kg
Total mass of the mixture = mass of A + mass of B + mass of C + mass of water
= 0.3 + 0.5 + 0.4 +9.8
= 11 kg
Moles of A = 0.3/50 = 0.006 kg.moles
Moles of B = 0.5/250 = 0.002 kg.moles
Moles of C = 0.4/10 = 0.04 kg.moles
Moles of water = 9.8/18 = 0.544 kg.moles
Total moles present = moles of A + moles of B + moles of C + moles of water
= 0.006 + 0.002 + 0.04 + 0.544
= 0.592 kg.moles
a)
Mass concentration of A = 0.3/10 = 0.03 kg/m3
Mass concentration of B = 0.5/10 = 0.05 kg/m3
Mass concentration of C = 0.4/10 = 0.04 kg/m3
Mass concentration of water = 9.8/10 = 0.98 kg/m3
b)
Mole concentration of A = 0.006/10 = 0.0006 kg.moles/m3
Mole concentration of B = 0.002/10 = 0.0002 kg.moles/m3
Mole concentration of C = 0.04/10 = 0.004 kg.moles/m3
Mole concentration of water = 0.544/10 = 0.0544 kg.moles/m3
c)
Total mole concentration = mole concentration of A + mole concentration of B + mole concentration of C+ mole concentration of water
= 0.006 + 0.002 + 0.004 + 0.0544
= 0.0644 kg.moles/m3
Total mass concentration = mass concentration of A + mass concentration of B + mass concentration of C+ mass concentration of water
= 0.03 + 0.05 + 0.04 + 0.98
= 1.1 kg/m3
d)
Mass fraction of A = 0.3/11 = 0.0273
Mass fraction of B = 0.5/11 = 0.04545
Mass fraction of C = 0.4/11 = 0.03636
Mass fraction of water = 9.8/11 = 0.891
Mole fraction of A = 0.006/0.592 = 0.010135
Mole fraction of B = 0.002/0.592 = 0.003378
Mole fraction of C = 0.04/0.592 = 0.6757
Mole fraction of water = 0.544/0.592 = 0.91892