A mixture of CaCl_(2) and NaCl weighing 4.44 g is treated with sodium carbonate solution to precipitate all the Ca^(2+) 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 (atomic mass of Ca = 40) is

A mixture of CaCl_(2) and NaCl weighing 4.44 g is treated with sodium

1.	A mixture of CaCl_(2) and NaCl weighing 4.44 g is treated with sodium carbonate solution to precipitate all the Ca^(2+) 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 (atomic mass of Ca = 40) is
Answer» 75 30.6 25 69.4 Solution :Suppose `CaCl_(2)` in the mixture = x G `underset(111g)(CaCl_(2))+Na_(2)CO_(3)RARR underset(100g)(CaCO_(3))+2NaCl` 111 g `CaCl_(2)` PRODUCE `CaCO_(3)=100g` `therefore"x g "CaCl_(2)" will produce "CaCO_(3)=(100)/(111)xg` `underset(100g)(CaCO_(3))rarrunderset(56g)(CaO)+CO_(2)` 100 g `CaCO_(3)` produce `CaO=56g` `therefore (100x)/(111)g" will produce CaO"=(56)/(100)xx(100x)/(111)` `=(56x)/(111)g` Thus, `(56x)/(111)=0.56"or"x=1.11 g` `therefore %" of NACL in the mixture "=(3.33)/(4.44)xx100=755` Alternatively, `underset(111g)(CaCl_(2))rarrCaCO_(3)rarrunderset(56g)(CaO)` x g `CaCl_(2)` will GIVE `CaO=(56x)/(111)g` `therefore""(56x)/(111)=0.56"(given)or"x=1.11g`