Question

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:

V_mix= 10 m³

m_A= 0.3 kg , m_B= 0.5 kg , m_C= 0.4 kg , m_H2O= 9.8 kg

MW_A= 50 , MW_B= 250 , MW_C= 10 , MW_H2O= 18

Answer 1

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