A mixture of CaCl2 and NaCl weighing 4.44 g is treated with sodium carbonate solution to precipitate all the Ca2+ ions as calcium carbonate. The calcium carbonate so obtained is heated strongly to get 0.56 g of CaO. The percentage of NaCl in the mixture is (Atomic mass of Ca = 40)

CaCO3⟶ΔCaO+CO2 1 mol          1 molCaCl21 mol+Na2CO3⟶CaCO31 mol+2NaCl1 mol of CaO≅1 mol of CaCl20.5656 mol of CaO≅0.01 mol of CaCl2                            =0.01×111 g CaCl2                             =1.11 g CaCl2Thus, in the mixture, weight of NaCl = 4.44- 1.11= 3.33 g∴   Percentage of NaCl=3.334.44×100=75%