CBSE Class 11-science Answered

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 = ?

N_HCl × V_HCl = N_CaCO3 × V_CaCO3

V_HCl = 50/0.1 = 500 cm³