2. When 1 litre of 1 M NaCl solution in water is mixed with 1 litre of 1M CaCl2 solution
in water, what is the total concentration of chloride ions in the resulting solution?
a. 2 M
b. 1 M
c. 3 M
d. 1.5 M
Problem Solution
Mixing 1 litre of 1 M NaCl with 1 litre of 1 M CaCl₂ produces a solution with total chloride ion concentration of 1.5 M. NaCl dissociates into 1 Na⁺ and 1 Cl⁻, yielding 1 mole of Cl⁻ from 1 mole of NaCl. CaCl₂ dissociates into 1 Ca²⁺ and 2 Cl⁻, yielding 2 moles of Cl⁻ from 1 mole of CaCl₂.[execute_python]
Total moles of Cl⁻ equal 1 (from NaCl) + 2 (from CaCl₂) = 3 moles in 2 litres total volume. Thus, concentration = 3 moles / 2 L = 1.5 M, corresponding to option d.
Option Analysis
Option a: 2 M
This assumes average of 1 Cl⁻ from NaCl and 2 Cl⁻ from CaCl₂ without volume adjustment, ignoring dilution effect on total moles over 2 L. Incorrect as it undercounts Cl⁻ contribution.
Option b: 1 M
This treats solutions as additive by molarity without mole calculation, as if each contributes 1 M Cl⁻ equally. Fails to account for CaCl₂’s double Cl⁻ ions.
Option c: 3 M
This sums molarities (1 M + 2 M from CaCl₂) without dividing by doubled volume. Overlooks dilution principle where moles remain constant but volume doubles.
Option d: 1.5 M (Correct)
Matches calculation: total 3 moles Cl⁻ in 2 L. Formula M_total × V_total = Σ (M_i × V_i × ν_Cl), where ν_Cl is Cl⁻ per formula unit (1 for NaCl, 2 for CaCl₂).[execute_python]
Key Takeaway for CSIR NET
This question tests mole conservation and dissociation in mixtures, common in solution chemistry sections. Practice similar dilutions using M1V1 = M2V2 adapted for ions.
