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. 75
B. 30.5
C. 25
D. 69.4

Answer

A

Explanation

Concerned reaction is
`underset(100 g)(CaCO_(3)) overset(Delta)(rarr) underset(56 g)(CaO)+CO_(2)`
56 g of CaO is obtained from 100 g of `CaCO_(3)`
0.56 g of CaO is obtained by `100/56xx0.56=1 g` of `CaCO_(3)` we know that,
`underset(111 g)(CaCl_(2))+Na_(2)CO_(3) rarr underset(100 g)(CaCO_(3))+2 NaCl`
100 g of `CaCO_(3)` is obtained by 11 g of `CaCl_(2)`
1 g of `CaCO_(3)` is obtained by `111/100=1.11 g` of `CaCl_(2)`
Weight of `NaCl=4.44-1.11=3.33 g`
% age of `NaCl=3.33/4.44xx100=75%`.