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A mixture of CaCl2 and NaCl weighing 4...

A mixture of `CaCl_2` and NaCl weighing 4.44g 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 0.56g of CaO. The percentage of NaCl in the mixture is [Atomic mass of Ca = 40]

A

31.5

B

75

C

25

D

40.2

Text Solution

Verified by Experts

The correct Answer is:
B
`NaCl+ CaCl_2 overset(Na_2CO_3)(to) CaCO_3`
Let wt. of `CaCl_2` in the mixture = x and wt of NaCl in the mixture = 4.44 - x
` CaCO_3 overset(Delta)(to) CaO + CO_2`
moles of Ca in `CaO_2` = moles of Ca in `CaCO_3` = moles of Ca in CaO
` therefore (x)/(111) = (0.56)/(56)`
x = 1.11 g
wt. of NaCl in mixture = 4.44 - 1.11 = 3.33 g
% of NaCl` = (3.33)/(4.44) xx 100 = 75%`
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