A mixture of `CaCl_(2)` and NaCl weighing 4.44 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 of (atomic mass of Ca=40) is

Answer

75

Explanation

`underset(111 g)(CaCl_(2)) + Na_(2)CO_(3) to underset(100 g)(CaCO_(3)) + 2 NaCl`
Let the x g of `CaCl_(2)` in the mixture.
`therefore 111 g CaCl_(2)` gives = 100 g `CaCO_(3)`
`therefore x g CaCl_(2)` will give `=(100 xx x)/111 g CaCO_(3)`
Now `underset(100 g)(CaCO_(3)) to underset(56 g)(CaOH) + CO_(2)`
`therefore` 100 g `CaCO_(3)` gives = 56 g CaO
`therefore (100 xx x)/111 g CaCO_(3)`, will give `= (100 xx x xx 56)/(100 xx 111)`
`= (56 x)/111`g CaO
`therefore (56 x)/111 =0.56, therefore x = 1.11 g`
`therefore` Mass of NaCl in mixture `=4.44 - 1.11 =3.33 g`
% NaCl in mixture `= (3.33 xx 100)/4.44 = 75%`