Question

A solution is 5.43 % by mass benzene (C_6H_6) in carbon tetrachloride (CCl_4). 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 mole fraction, molarity and molality of the benzene?

Answer 1

Given

Mass % of benzene is 5.43 = (mass of benzene / mass of solution ) * 100 %

Let us take a basis of 100 g of solution

so mass of benzene in 100 g of solution will be

5.43 % = (mass of benzene / 100 g ) * 100 %

Mass of benzene = 5.43 g

Mass of CCl4 (solvent) = Mass of solution - Mass of solute (benzene) = 100 g - 5.43 g = 94.57 g = 0.09457 kg

Given density of solution = 1.55 g/ml

Volume of solution = Mass of solution / density = 100 g / 1.55 g/ml = 64.52 ml = 0.06452 L

Molar mass of benzene = 78.1 g/mol

Molar mass of CCl4 = 154 g/mol

No. of moles of benzene = Mass of Benzene / Molar mass of benzene = 5.43 g / 78.1 g/mol = 0.0695 moles

No. of moles of CCl4 = Mass of CCl4 / Molar mass of CCl4 = 94.57 g / 154 g/mol = 0.6141 moles

total no. of moles = No. of moles of Benzne + No. of moles of CCl4 = 0.0695 + 0.6141 = 0.6836 moles

Mole fraction of benzene = No. of moles of benzene / total no. of moles = 0.0695 moles / 0.6836 moles = 0.1017

Mole fraction = 0.1017 ---------> Answer

Molarity of benzene = No. of moles of benzene / Volume of solution in L = 0.0695 moles / 0.06452 L = 1.08 mol/L

Molarity of benzene = 1.08 mol/L or M ----------> Answer

Molality of benzene = No. of moles of benzene / mass of solvent in kg = 0.0695 moles / 0.09457 kg = 0.735 mol/kg

Molality of benzene = 0.735 mol/kg or m ---------> Answer