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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) isA. 75B. 30.5C. 25D. 69.4 |
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Answer» Correct Answer - A 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%`. |
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