Electrolytes - 2 | Chemical Calculations for Solutions

1) Copper(II) chloride: CuCl₂ (s)

CuCl₂ (s) + H₂O (l) → Cu²⁺ (aq) + 2 Cl^- (aq)

0.250 M copper(II) chloride ⇒ 0.250M of Cu²⁺ and 0.500M of Cl^-

Moles of Cu²⁺: 

n_Cu2+ (in mol) = M_Cu2+ (in mol.L^-1)  x V (in L)
n_Cu2+ = 0.250 x 50.0 x 10^-3 = 1.25 x 10^-2 mol

Moles of Cl^-: 

n_Cl- (in mol) = M_Cl- (in mol.L^-1)  x V (in L)
n_Cl- = 0.500 x 50.0 x 10^-3 = 2.50 x 10^-2 mol

Total number of moles = (1.25 + 2.50) x 10^-2 = 3.75 x 10^-2 moles

2) I = $\frac{1}{2}$ $\sum _{i = 1}^n c_i {z_i}^2$

I = $\frac{1}{2}$ ( M_Cu2+ x 2² + M_Cl- x (-1)² ) = $\frac{1}{2}$ (0.250 x 4 + 0.500) = 0.75M