Question

A solution has a 0.102 mole fraction of benzene (C₆H₆) in carbon tetrachloride (CCl₄). The density of the solution is 1.55 g/mL. The molar masses of benzene and carbon tetrachloride are 78.1 g/mol and 154 g/mol respectively. What is the mass percent, molarity and molality of the benzene?

Answer 1

Mole fraction of benzene (C₆H₆) in the solution = 0.102; mole fraction of carbon tetrachloride (CCl₄) = (1 – 0.102) = 0.898 (sum of mole fractions is equal to 1).

Since the system contains only two components and the sum of the mole fractions is equal to 1.000, hence the mole fraction of C₆H₆ = mole of C₆H₆ and mole fraction of CCl₄ = mole of CCl₄.

Therefore, the solution contains 0.102 mole C₆H₆ and 0.898 mole CCl₄.

Mass of C₆H₆ = (mole of C₆H₆)*(molar mass of C₆H₆) = (0.102 mole)*(78.1 g/mol) = 7.9662 g.

Mass of CCl₄ = (0.898 mole)*(154 g/mol) = 138.292 g.

Total mass of the solution = (7.9662 + 138.292) g = 146.2582 g.

Mass percent of C₆H₆ =(mass of C₆H₆)/(mass of solution)*100% = (7.9662 g)/(146.2582 g)*100% = 5.4467% ≈ 5.45% (ans).

Volume of the solution = (mass of the solution)/(density of the solution) = (146.2582 g)/(1.55 g/mL) = 94.3601 mL = (94.3601 mL)*(1 L/1000 mL) = 0.09436 L.

Molarity of benzene =(mole of benzene)/(volume of solution in L) = (0.102 mole)/(0.09436 L) = 1.08097 mol/L ≈ 1.08 M (ans).

Mass of solvent (CCl₄) = 146.2582 g = (146.2582 g)*(1 kg/1000 g) = 0.1462582 kg.

Molality of benzene = (mole of benzene)/(kg of CCl₄) = (0.102 mole)/(0.1462582 kg) = 0.6973 m ≈ 0.70 m (ans).