Saved Bookmarks
| 1. |
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. `75`B. `30.6`C. `25`D. `69.4` |
|
Answer» Correct Answer - A `NaCl+Na_(2)CO_(3)to` No reaction `CaCl_(2)+Na_(2)CO_(3)toCaCO_(3)+2NaCl` `underset(=111g)((40+2xx35.5))underset(=106g)((23xx2+12+16xx3))underset(=100g)((40+12+3xx16))` Let the weight of `CaCl_(2)` in the mixture be x g `:.` weight of `NaCl` in the mixture `=(4.44-x)g` `CaCO_(3)overset(Delta)(to)CaO+CO_(2)` `40+16` `=56g` Now 111 g `CaCl_(2)` produce `CaCO_(3)=100g` `:.` x g of `CaCO_(2)` produce CaO = 56 g `(100)/(111)xCaCO_(3)` produce CaO `= (56)/(100)xx(100)/(111)xg` `=(56)/(111)xg=0.50g` or x = 1.11 g `:.` NaCl in the mixture = 4.44-1.11 = 3.33 g Percentage of NaCl in the mixture `=(3.33)/(4.44)xx100=75%` |
|