Question

100 cm³ of concentrated hydrochloric acid which is 11 M are diluted to 275 cm³. If 2.5 g of limestone are required to neutralize 10 cm³ of this diluted - acid, what is the percentage of calcium carbonate in the sample of limestone?

Answer 1

Substituting we get, 100 × 11 = 275 × M₂

∴ Normality of the diluted acid = 4 

1000 cm³ of the acid solution contain 4 gram 

∴ 10 cm³ of the diluted acid contain = 4 x 10/1000 i.e., 0.4 gram 

2.5 g of limestone contain 0.04 gram of calcium carbonate i.e., they contain 0.04 × 100 g = 4 g of calcium carbonate. 

(Molar mass of calcium carbonate = 100) 

∴ percentage of calcium carbonate in the sample of limestone = 4 x 100/2.5 = 16.0.