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` in 50 mL of 0.5 M solution
` = (0.5)/(100) xx 50 = 2.5 xx 10^(-2) ` mol
` therefore ` moles 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 xx 10^(-2) `mol
volume of 0.1N HCl required x, is calculated using the relation .
molarity = no. of moles/vol. of solution (mL) x 1000
`0.1 HCl = 0.1 M HCl = (5 xx 10^(-2) )/(x) xx 1000`
` therefore x = (5 xx 10^(-2) xx 1000)/(0.1) `
` = 500 mL = 500 cm^3`