Question

A solution of NaCl (MW = 58.5 g/mol) in water (MW = 18.0 g/mol) has the molality of 3.01 molal and the density of 1.112 g/mL. Consider mass of solvent 1 Kg A) What is the mass percent of the solute? (B) What is the molarity of the solution? (C)What is the mole fraction of H2O?

Answer 1

molality = moles of solute (NaCl)/ kg of solvent =3.01

Basis : 1 kg of water solvent, moles of solvent =3.01, mass of NaCl = moles* Molecular weight = 3.01*58.5=176.085 gm

Moles of water = mass of water / Molecular weight of water =1000/18=55.56 moles and moles of NaCl= 3.01

Mole fraction of H2O= moles of water/ Total moles = 55.56/(55.56+3.01)=0.948

Total mass= 1000gm ( solvent) + 176.085 (NaCl)= 1176.085 gm

Mass % of solute = 100*176.085/1176.085=14.97%

Density = 1.112 g/ml, volume= mass/ density

Volume= 1176.085/1.112=1057.63 ml= 1.0576 L

1.0576 L contains 3.01 moles of NaCl

1L contains 3.01/1.0576 =2.85 M ( this is molarity)