The total number of pair of compounds, in which $1^{st}$ compound is more soluble than $2^{nd}$ compound is

(BeSO₄ & BaSO₄), (BeF₂ & BaF₂), (Na₂CO₃ & Cs₂CO₃), (CaCrO₄ & BaCrO₄), (LiCl & KCl), (NaHCO₃ & KHCO₃)

Question

The total number of pair of compounds, in which $1^{st}$ compound is more soluble than $2^{nd}$ compound is

(BeSO₄ & BaSO₄), (BeF₂ & BaF₂), (Na₂CO₃ & Cs₂CO₃), (CaCrO₄ & BaCrO₄), (LiCl & KCl), (NaHCO₃ & KHCO₃)

TestHub · Accepted Answer

Compare solubility of given pairs: BeSO₄ > BaSO₄, BeF₂ > BaF₂, Na₂CO₃ > Cs₂CO₃, LiCl > KCl → total 4 pairs

Chemistry - Ionic Equilibrium Question with Solution | TestHub