Express of `CO_(2)` 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 carbonated was completely neutralized with 0.1 N hydrochloric acid. The volume of hydrochloric acid required is (At mass of carbon = 40)

Options

A. `200ml`
B. `500ml`
C. `400ml`
D. `300ml`

Answer

B

Explanation

According to the question, the reaction occurs as
`underset(74(=1mol))(Ca(OH)_(2))+underset(44(=1mol))(CO_(2))to underset(100(=1mol))(CaCO_(3))+underset(18(=1mol))(H_(2)O)`
Given, 50 ml of 0.5 M `Ca(OH)_(2)` reacts with excess of `CO_(2)`
`therefore` No. of millimoles of `Ca(OH)_(2)` reacts with excess of `CO_(2)`
`therefore` No. of millimoles of `Ca(HO)_(2)` reacted=25
`therefore` 1 mole of `Ca(OH)_(2)` gives 1 mole of `CaCO_(3)`,
`therefore`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)=1xx50impliesV_(HCl)=(50)/(0.1)=500cm^(3)`.