A mixture of CaCl_2and NaCl weighing 4.44gis treated with sodium carbonate solution to precipitate all the calcium ions as calcium carbonate. The calcium carbonate so obtained is heated strongly to get 0.56g of CaO. The percentage of NaCl in the mixture is [Atomic mass of Ca = 40]

A mixture of CaCl_2and NaCl weighing 4.44gis treated with sodium carbo

1.	A mixture of CaCl_2and NaCl weighing 4.44gis treated with sodium carbonate solution to precipitate all the calcium ions as calcium carbonate. The calcium carbonate so obtained is heated strongly to get 0.56g of CaO. The percentage of NaCl in the mixture is [Atomic mass of Ca = 40]
Answer» 31.5 75 25 40.2 Solution :GIVEN that all the calcium PRESENT in the miture will convert into its carbonates and again on heating it gives its OXIDE CaO of weight 0.56 g. Moles of CA in `CaCl_(2)` =moles of Ca in `CaCO_(3)` Moles of Ca in `CaCO_(3)`=moles of Ca in CaO `implies(X)/(111)=(0.56)/(56)impliesx(0.56)/(56)xx111=1.11g` [m.wt. of `CaCl_(2)`=111,m.wt. of CaO=56] We know that `NaCl+CaCl_(2)`=4.44 y+1.11=4.44`implies`y=3.33g Percentage of NaCl=`3.33xx(100)/(4.44)`=75%