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

Excess of carbon dioxide is passed through 50 mL of 0.5 M calcium hydr

1.	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
Answer» `200 cm^(3)` `500 cm^(3)` `400 cm^(3)` `300 cm^(3)` Solution :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)`