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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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$300 {cm}^{3}$

$200 {cm}^{3}$

$500 {cm}^{3}$

$400 {cm}^{3}$

answer is C.

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Detailed Solution

$Number of millimoles of Ca (OH)_{2} = 50 \times 0.5 = 25$

$Number of millimoles of {CaCO}_{3} = 25$

$Number of milliequivalents of {CaCO}_{3} = 50$

$∴ Volume of 0.1 NHCl = \frac{50}{0.1} = 500 {cm}^{3}$

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