Q. A mixture of ${CaCl}_{2}$ and NaCl weighing 4.44 g 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 (atomic mass of $Ca = 40$) is
A) 75
B) 30.6
C) 25
D) 69.4
Explanation
(a)
$CaCO_{3}\overset{\mathrm{\Delta}}{\rightarrow}CaO + CO_{2}$
1 mol 1 mol
$CaCl_{2} + Na_{2}CO_{3} \rightarrow CaCO_{3} + 2Na_{2}$
1 mol 1 mol
1 mol $CaO \cong$1 mol $CaCl_{2}$
$\frac{0.56}{56}$ mol CaO$\cong$0.01 mol $CaCl_{2}$
$= 0.01 \times 111$ g$CaCl_{2}$
$= 1.11\ g\ CaCl_{2}$
Thus, in the mixture, weight of
NaCl=4.44$-$1.11=3.33 g
$\therefore$ Percentage of NaCl$= \frac{3.33}{4.44} \times 100 = 75\%$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.