Question

2M of 100 ml Na₂SO₄ is mixed with 3M of 100ml NaCl solution and 1M of 200 ml CaCl₂ solution. Then the ratio of the concentration of cation and anion

A) 1/2

B) 2

C) 1.5

D) 1

Answer 1

Answer : Option D is correct

Solution 1 : Na2SO4
Na2SO4 → 2Na+ + SO42-
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 SO42- = 2 M in 100 ml
Number of moles of SO42- = Molarity x Volume (L) = 2 x (100/1000) = 0.2 mol of SO42-


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