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

According to the question the reaction occurs as
`underset((=1"mol"))underset(74)(Ca(OH)_(2))+underset((=1"mol"))underset(44)(CO_(2))tounderset((=1"mol"))underset(100)(CaCO_(3))+underset((=1"mol"))underset(18)underset(H_(2)O`
Given 50 ml of 0.5 M `Ca(OH)_(2)` reacts with excess of `CO_(2)`
`:.` no. of millimoles of `Ca(OH)_(2)` reacted =25
`:'` 1 mile of `Ca(OH)_(2)` gives 1 mile of `CaCO_(3)`,
`:.` no. of millimoles of `CaCO_(3)` formed =25
`:.` no. of millieq `=("weight in mg")/("eq. wt")=(25xx100)/50=50`
`implies` no. of milliequivalents of `CaCO_(3)=50`
As volume of `CaCO_(3)` solution =50 ml
So, normality of `CaCO_(3)` solution =1N
(millieq `=NxxV` in ml)
Normality of HCl `=0.1N` (given)
Volume of HCl=?
`N_(HCl)xxV_(HCl)=N_(CaCO_(3))xxV_(CaCO_(3))`
`implies0.1xxV_(HCl)=1xx50`
`impliesV_(HCl)=50/0.1=00 cm^(3)`