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)

Question

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) 500 cm3 (B) 400 cm3 (C) 300 cm3 (D) 200 cm3

Tardigrade Admin · Accepted Answer

No. of millimoles of Ca ( OH )2=50 × 0.5=25 No. of millimoles of CaCO 3=25 No. of milliequivalents of CaCO 3=50 ∴ Volume of 0.1 N HCl =(50/0.1)=500 cm 3

$500 c m^{3}$

$400 c m^{3}$

$300 c m^{3}$

$200 c m^{3}$

No. of millimoles of $C a (O H)_{2} = 50 \times 0.5 = 25$

No. of millimoles of $C a C O_{3} = 25$

No. of milliequivalents of $C a C O_{3} = 50$

$∴$ Volume of $0.1 N H Cl = \frac{50}{0.1} = 500 c m^{3}$

500cm3

400cm3

300cm3

200cm3

No. of millimoles of Ca(OH)2​=50×0.5=25 No. of millimoles of CaCO3​=25 No. of milliequivalents of CaCO3​=50 ∴ Volume of 0.1NHCl=0.150​=500cm3

$500 c m^{3}$

$400 c m^{3}$

$300 c m^{3}$

$200 c m^{3}$

No. of millimoles of $C a (O H)_{2} = 50 \times 0.5 = 25$

No. of millimoles of $C a C O_{3} = 25$

No. of milliequivalents of $C a C O_{3} = 50$

$∴$ Volume of $0.1 N H Cl = \frac{50}{0.1} = 500 c m^{3}$