Answer:
(a) CS₂ = 0.331, CCl₄ = 0.669; (b) 102 g/mol; (c) 49.1 kmol/h
Explanation:
Assume 5000 kg of mixture.
(a) Mole fractions
(i) Mass of each compound
Mass of CCl₄ = 2500 kg
Mass of CS₂ = 2500 kg
(ii) Moles of each compound
[tex]\text{Moles of CCl}_{4} = \text{2500 kg CCl}_{4} \times \dfrac{\text{1 kmol CCl}_{4}}{\text{153.82 kg CCl}_{4}} = \text{16.3 kmol CCl}_{4}\\\\\text{Moles of CS}_{2} = \text{2500 kg CS}_{2} \times \dfrac{\text{1 kmol CS}_{2}}{\text{76.14 kg CS}_{2}} = \text{32.8 kmol CS}_{2}[/tex]
(iii) Mole fractions
Total moles = 16.3 kmol + 32.8 kmol = 49.1 kmol
The major component is usually called the solvent — Component 1 —carbon disulfide.
[tex]\chi_{2} = \dfrac{n_{2}}{n_{\text{tot}}} =\dfrac{\text{16.3}}{\text{49.1}} = \mathbf{0.331}\\\\\chi_{1} = \dfrac{n_{1}}{n_{\text{tot}}} = \dfrac{\text{35.8}}{\text{49.1}} = \mathbf{0.669}[/tex]
(c) Average molar mass
The average molar mass is the weighted molar mass of each component.
[tex]\text{MM} =\chi_{2} \text{M}_{2} + \chi_{1} \text{M}_{1} = 0.331 \times \dfrac{\text{153.82 g }}{\text{1 mol}} + 0.669 \times \dfrac{\text{76.14 g}}{\text{1 mol}} = \dfrac{\text{50.9 g}}{\text{1 mol }} + \dfrac{\text{50.9 g}}{\text{1 mol }}\\\\=\textbf{102 g/mol}[/tex]
(d) Feed rate
[tex]\text{Feed rate} = \dfrac{\text{total moles}}{\text{time}} = \textbf{49.1 kmol/h}[/tex]
