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 (At. mass of carbon = 40)

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 (At. mass of carbon = 40)
Answer» 200 mL 500 mL 400 mL 300 mL Solution :`Ca(OH)_(2)+CO_(2)RARR CaCO_(3)+H_(2)O` 50 mL of 0.5 `Ca(OH)_(2)` contain `Ca(OH)_(2)` `=(0.5)/(1000)xx"50 mol"=0.025"mol"` 1 mol of `Ca(OH)_(2)` PRODUCES `CaCO_(3)` = 1 mol `therefore 0.25` mole of `Ca(OH)_(2)` will produce `CaO_(3)` = 0.025 mol `CaCO_(3)+2HClrarr CaCl_(2)+H_(2)O+CO_(2)` 1 mole of `CaCO_(3)` is NEUTRALISED by HCL = 2 mol `therefore 0.025` mol of `CaCO_(3)` will be neutralised by HCl `=2xx0.025" mol" = 0.05 " mol"` `" 1000 mL of 0.1 N HCl contain HCl = 0.1 g eq"` `"= 0.1 mol"` `"or 0.1 of HCl is present in HCl sol = 1000 mL"` `therefore" 0.05 mol of HCl will be present in HCl sol."` `=(1000)/(0.1)xx0.05mL="500 mL"`