KCET · Chemistry · Some Basic Concepts of Chemistry
Excess of carbon dioxide is passed through 50 \(\mathrm{mL}\) of \(0.5 \mathrm{M}\) calcium hydroxide solution. After the completion of the reaction, the solution was evaporated to dryness. The solid calcium carbonate was completely neutralised with \(0.1 \mathrm{~N}\) hydrochloric acid. The volume of hydrochloric acid required is (Atomic mass of calcium \(=40\) )
- A \(300 \mathrm{~cm}^{3}\)
- B \(200 \mathrm{~cm}^{3}\)
- C \(500 \mathrm{~cm}^{3}\)
- D \(400 \mathrm{~cm}^{3}\)
Answer & Solution
Correct Answer
(C) \(500 \mathrm{~cm}^{3}\)
Step-by-step Solution
Detailed explanation
No. of millimoles of \(\mathrm{Ca}(\mathrm{OH})_{2}=50 \times 0.5=25\)
No. of millimoles of \(\mathrm{CaCO}_{3}=25\)
No. of milliequivalents of \(\mathrm{CaCO}_{3}=50\)
\(\therefore\) Volume of \(0.1 \mathrm{~N} \mathrm{HCl}=\frac{50}{0.1}=500 \mathrm{~cm}^{3}\)
No. of millimoles of \(\mathrm{CaCO}_{3}=25\)
No. of milliequivalents of \(\mathrm{CaCO}_{3}=50\)
\(\therefore\) Volume of \(0.1 \mathrm{~N} \mathrm{HCl}=\frac{50}{0.1}=500 \mathrm{~cm}^{3}\)
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