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

Options

A. 31.5
B. 75
C. 25
D. 40.2

Answer

B

Explanation

`NaCl + CaCl_(2) overset(Na_(2)CO_(3)) to CaCO_(3)`
let wt. of `CaCl_(2)` in mixture =x
and wt. of NaCl in mixture = 4.44 -x
`CaCO_(3) to CaO + CO_(2)`
Moles of Ca in `CaCl_(2)` = Moles of Ca in `CaCO_(3)`
Moles of Ca in `CaCO_(3)` = Moles of Ca in CaO
`x/111 = 0.56/56`
`therefore x =0.56/56 xx 111 = 1.11`g
Wt. of NaCl in mixutre `=4.44 -1.11 = 3.33`g
% of NaCl `=3.33/4.44 xx 100 = 75%`