Question

An aqueous solution containing sodium chloride (NaCl) and calcium chloride (CaCl2) is found to have a density of 1.043 g/mL. The results of an analysis indicate that the calcium concentration is 8.3 g/L and the sodium concentration is 12.2 g/L. What is the mole fraction of each component in this solution?

Molar mass of Na = 22.99 g/mol

Molar mass of Cl = 35.45 g/mol

Molar mass of Ca = 40.078 g/mol

Answer 1

Basis : Volume of solution = 100ml, density of the solution = 1.043 gml, mass of the solution = density* volume= 1.043*100=104.3 gm

Molecular formula of sodium chloride= NaCl, molecular formula for Calcium chloride= CaCl2

NaCl contains 1 mole of sodium and 1 mole of chlorine and CaCl2 contains 1 mole of Ca and 2 moles of chlorine

given sodium concentration is 12.2 g/L. volume of solution = 100ml, mass of sodium in 100ml= 12.2*100/1000=1.22 gm, moles of sodium= mass/molar mass = 1.22/22.99 = 0.053 , moles of Cl = 0.053, mass of CaCl2 =8.3*100/1000=0.83 gm,, moles of Calcium= 0.83/40 .078=0.02075, moles of Cl2= 2* moles of Calciium= 2*0.02075=0.0415

mass of Chlorine from NaCl= molar mass of Cl* moles of Cl from NaCl= 0.053*35.545= 1.88gm , mass of Cl from CaCl2= 0.0415*35.45= 1.47 gm

mass of NaCl= mass of Na+ mass of Cl =1.88+1.22=3.1 gm, mass of CaCl2= 1.47+0.83= 2.3 gm

mass of the solution= 104.3 gm, mass of water= total mass of the solution-mass of NaCl-mass of CaCl2= 104.3-3.1-2.3= 98.9 gm

masses (gm); NaCl= 3.1, CaCl2= 2.3 and water= 98.9

molar masses : NaCl= 22.99+35.45=58.44, CaCl2= 40.078+2*35.45= 110.978, water= 18

moles= mass/molar mass

moles : NaCl= 3.1/58.44=0.053, CaCl2= 2.3/110.978=0.021 and water= 98.9/18= 5.49

total moes= moles of NaCl+ moles of CaCl2+ moles of water= 0.053+0.021+5.49=5.564

mole fraction= moles/ total moles

Mole fractions: NaCl= 0.053/5.564= 0.009, CaCl2= 0.003 and water= 1-0.003-0.009=0.988