Question

Which of the two aqueous solutions will be a better conductor of electricity in the following examples? Explain why in each case.
a)1.0M NaCl vs. 0.010 M Nacl
b)0.10 M NaCl vs. 0.10 M Al2(SO4)3 (assume complete dissociation of both salts).
c) 1.0 mole of NaCl added to water vs. 1.0 mole of pbCl2 added to water (hint: remember the solubility rules).

Answer 1

a) The solution with a concentration of 1.0 M NaCl will be a better conductor of electricity since it has a higher number of electrolytes
b) The 0.10 M Al2(SO4)3 will be a better conductor since more ions are formed per mole of the salt dissociated
c) 1.0 mol of NaCl in water because NaCl is very soluble in water

Answer 2

Aqueous solutions that contain ionic compounds can conduct electricity. Greater the number of ions in the solution, better will be the electrical conductivity.

a) 1.0 M NaCl dissociates to give $1 mol Na^{+}$and $1 mol Cl^{-}$ions per liter of the solution. So, it would produce 2 mol ions. Whereas as 0.010 M NaCl can produce$0.010 mol Na^{+}$and [tex0.010 mol Cl^{-}[/tex]ions per liter of the solution, that is a total of 0.020 mol ions.Hence, 1.0 M NaCl solution will be a better conductor of electricity.

b) 0.10 M NaCl can produce$0.10 mol Na^{+}$and [tex0.10 mol Cl^{-}[/tex]ions per liter of the solution, that is a total of 0.20 mol ions. 0.10 M $Al_{2}(SO_{4})_{3}$produces 0.20 mol$Al^{3+}$ions and 0.30 mol$SO_{4}^{2-}$ions, that is around 0.50 mol ions. So, 0.10 M $Al_{2}(SO_{4})_{3}$will be a better conductor.                          

       c) NaCl is completely soluble in water so it can ionize completely. $PbCl_{2}$is sparingly soluble in water, so it cannot ionize completely. Therefore, 1.0 mol NaCl will be a better conductor than 1.0 mol PbCl$_{2}$