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

Options

A. A) 75
B. B) `30.6`
C. C) 25
D. D) `69.4`

Answer

A

Explanation

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