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. 30.6
C. 25
D. 69.4

Answer

A

Explanation

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`