Question

A mixture of CaCl2, and NaCl weighing 4.44 g is treated with sodium carbonate solution to precipitate all the calcium ions as calcium carbonate. The calcium carbonate so obtained is heated strongly to get 0.56 g of CaO. The percentage of NaCl in the mixture is [Atomic mass of Ca = 40]

Answer 1

Answer:

Explanation:

CaCl2 + NaCl + NaC03 → CaC03

4.44     3.44

Co.C03milli  = CaCO3milli

x = 3.44

%purity= 3.44/4.44 x 100

=75%

Answer 2

Answer:

75

Explanation:

Weight of CaO = 0.56 g

0.56 g of CaO contains 0.40 g of calcium. Mol. weight of CaCl2 = 40 + 71 = 111

40 g of calcium present in 111 g of CaCl2

∴ 0.4 g of calcium present in 0.40 x 111/40 = 1.11 g of CaCl2

Amount of NaCl present in mixture = 4.44 – 1.11 = 3.33 g

∴ Percentage of NaCl = 3.33 /4.44x 100 = 75