Question

Importance of basis: A liquid salt solution contains 5%(by mass)NaCl,7%(by mass)MgCl2)3% by mass)KCl and 85%(by mass)water(H2O). What are the average molecular weight of the solution(salt plus water). What is the average molecular weight of the water free of solution( that is , the average molecular weight of the NaCL,MgCL2, and KCl components of the solution)? Finally, if the specific gravity of the solution (salt and water) is 1.14, what are the concentrations of each of the salts(NaCl,MgCl2,KCl) in g/L

Answer 1

Part a

Basis - 100 g of salt solution

Moles of NaCl = mass/molecular weight

= 5/58.4 = 0.0856 moles

Moles of MgCl2 = 7/95.21 = 0.0735

Moles of KCl = 3/74.55 = 0.0402

Moles of H2O = 85/18 = 4.722

Total Moles = 0.0856 + 0.0735 + 0.0402 + 4.722

= 4.9213

Mol fraction of NaCl X1 = moles of NaCl/total moles

= 0.0856/4.9213 = 0.01739

Mol fraction of MgCl2 X2 = moles of MgCl2/total moles

= 0.0735/4.9213 = 0.01493

Mol fraction of KCl X3= moles of KCl/total moles

= 0.0402/4.9213 = 0.00816

Mol fraction of H2O X4= moles of H2O/total moles

= 4.722/4.9213 = 0.9595

Average molecular weight of salt

= M1*X1 + M2*X2 + M3*X3 + M4*X4

= 58.4*0.01739 + 95.21*0.01493 +74.55*0.00816 + 18*0.9595

= 20.3163

Part b

Total Moles = 0.0856 + 0.0735 + 0.0402

= 0.1993

Mol fraction of NaCl X1 = moles of NaCl/total moles

= 0.0856/0.1993 = 0.4295

Mol fraction of MgCl2 X2 = moles of MgCl2/total moles

= 0.0735/0.1993 = 0.3687

Mol fraction of KCl X3= moles of KCl/total moles

= 0.0402/0.1993 = 0.2017

average molecular weight of the water free of solution

= M1*X1 + M2*X2 + M3*X3

= 58.4*0.4295 + 95.21*0.3687 +74.55*0.2017

= 75.223

Part C

Density of solution = 1.14*1000 = 1140 g/L

Concentration of NaCl = 0.01739 * 1140 g/L = 19.8246 g/L

Concentration of MgCl2 = 0.01493 * 1140 g/L = 17.0202 g/L

Concentration of KCl = 0.00816 * 1140 g/L = 9.3024 g/L

Concentration of H2O = 0.9595 * 1140 g/L = 1093.83 g/L