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

Options

A. `200 cm^3`
B. `500 cm^3`
C. `400 cm^3`
D. `300 cm^3`

Answer

B

Explanation

`Ca(OH)_2 + CO_2 to CaCO_3 + H_2O`
Moles of `Ca(OH)_2` present in 50 mL of 0.5 solution ` = (0.5)/(1000) xx 50 = 2.5 xx 10^(-2) ` mol
mass of `CaCO_3 ` formed ` = 2.5 xx 10^(-2) `mol
`CaCO_3 + 2HCl to CaCl_2 + H_2O + CO_2`
moles of HCl required ` = 2 xx 2.5 xx 10^(-2) ` mol
` = 5.0 xx 10^(-2) ` mol
volume of 0.1 N HCl required
` = (0.1)/(1000) xx x = 5.0 xx 10^(-2) `mol
` x = (5.0 xx 10^(-2) xx 1000)/(0.1) = 500 mL `