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]

Options

A. 31.5
B. 75
C. 25
D. 40.2

Answer

B

Explanation

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%