Q. Excess of carbon dioxide is passed through 50 mL of 0.5 M calcium hydroxide solution. After the completion of the reaction, the solution was evaporate was evaporated to dryness. The solid calcium carbonate was completely neutralised with 0.1 N hydrochloric acid. The volume of hydrochloric acid required is (Atomic mass of calcium=40)
A) $300\ {cm}^{3}$
B) $200\ {cm}^{3}$
C) $500\ {cm}^{3}$
D) $400\ {cm}^{3}$
Explanation
(c)
No. of Millimoles of $Ca(OH)_{2} = 50 \times 0.5 = 25$
No. of Millimoles of $CaCO_{3}$= 25
No. of milliequivalents of $CaCO_{3} = 50$
$\therefore$ Volume of 0.1 N HCl=$\frac{50}{0.1} = 500\ cm^{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.