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
A solution is 5.43 % by mass benzene (C_6H_6) in carbon tetrachloride (CCl_4). The density of...
A solution is prepared by dissolving 16.2 g of benzene (C6H6) in 282 g of 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, mole fraction, molarity and molality of the benzene?
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