Question

2 M of 100 ml sodium sulphate is mixed with 3 M of 100 ml sodium chloride solution and 1 M of 200 ml calcium chloride solution then the ratio of concentration of cation and anion is

Answer 1

Solution 1 : Na₂SO₄
Na₂SO₄ → 2Na⁺ + SO₄^2-
Molarity of Na₂SO₄​ = 2 M
Concentration of Na⁺ = 2 (2) M = 4 M in 100 ml
Number of moles of Na⁺ = Molarity x Volume (L) = 4 x (100/1000) = 0.4 mol of Na⁺

Concentration of SO₄^2- = 2 M in 100 ml
Number of moles of SO₄^2- = Molarity x Volume (L) = 2 x (100/1000) = 0.2 mol of SO₄^2-


Solution 2 : NaCl
NaCl → Na⁺ + Cl^-

Concentration of Na⁺ in 100 ml NaCl = 3 M in 100 ml
Number of moles of Na⁺ = Molarity x Volume (L) = 3 x (100/1000) = 0.3 mol of Na⁺

Concentration of Cl- in NaCl = 3 M in 100 ml
Number of moles = Molarity x Volume (L) = 3 x (100/1000) = 0.3 mol of Cl^-

Note-Actually calcium exist in the form of chloride as CaCl₂ but according to the given details of your question it is CaCl

CaCl₂ → Ca²⁺ + 2Cl^-

Concentration of Ca²⁺ = 1 M in 200 ml
Number of moles of Ca²⁺ = Molarity x Volume (L) =1 x (200/1000) = 0.2 mol of Ca²⁺

Concentration of Cl^- in 200 ml of CaCl₂ = 2 M in 200 ml
Number of moles of Cl^- = Molarity x Volume (L) = 2 x (200/1000) = 0.4 mol of Cl-

Therefore,

Total concentration of cation = 0.4 + 0.3 + 0.2 = 0.9 mol of Cation
Total concentration of anion= 0.2 + 0.3 + 0.4 = 0.9 mol of Anion

Answer-
Ratio of the concentration of cation and anion = 0.9/0.9 = 1