InterviewSolution
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InterviewSolution
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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 calcium=40)A. `200cm^(3)`B. `500cm^(3)`C. `400cm^(3)`D. `300cm^(3)` |
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Answer» Correct Answer - B 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)`. |
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