Excess of carbondioxide is passed through 50ml of 0.5M calcium hydroxide solution.After the completion of the reaction,the solution was evaporated to dryness.The solid calcium carbonate was completely neutralised with 0.1N Hydrochloric acid.The volume of hydrocholric acid required is(given atomic mass of calcium=40).

Asked by Malavika Umesh | 8th May, 2015, 04:29: PM

Expert Answer:

According to the given question, the chemical reaction can be written as follows:

Ca(OH)2 +  CO2 → CaCO3 + H2O

74               44            100               18

1 mole        1 mole       1 mole      1 mole

Now, 50 ml of 0.5 M Ca(OH)2 reacts with excess of CO2.

Number of millimoles of Ca(OH)2 reacted=25

1 mole  of Ca(OH)2 gives 1 mole of CaCO3.

No. of millimoles of CaCO3 formed=25

No. of milliequivalents of CaCO3 = 50

Since, volume of CaCO3 solution = 50 ml

Hence, normality of CaCO3 solution = 1 N

Normality of HCl=0.1 N

Volume of HCl = ?

NHCl × VHCl = NCaCO3 × VCaCO3

VHCl = 50/0.1 = 500 cm3

Answered by Prachi Sawant | 9th May, 2015, 11:11: PM