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 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)
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- $300~cm^{3}$
- $200~cm^{3}$
- $500~cm^{3}$
- $400~cm^{3}$
The Correct Option is C
Solution and Explanation
Millimoles of $Ca(OH)_{2} = 50 \times 0.5 = 25$.
Step 2: Stoichiometry
From $Ca(OH)_{2} + CO_{2} \rightarrow CaCO_{3}$, 25 millimoles of $Ca(OH)_{2}$ produce 25 millimoles of $CaCO_{3}$.
Step 3: Milliequivalents
$CaCO_{3}$ is bivalent ($n=2$), so milliequivalents $= 25 \times 2 = 50$.
Step 4: Acid Volume
Volume of 0.1 N HCl $= \frac{\text{Milliequivalents}}{\text{Normality}} = \frac{50}{0.1} = 500~cm^{3}$.
Final Answer: (c)
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