Ncert Exercises & Solutions

When water is added to compound (A) of calcium, solution of compound (B) is formed. When carbon dioxide is passes into the solution, it turns milky due to the formation of compound (C). If excess of carbon dioxide is passed into the solution milkiness disappears due to the formation of compound (D). Identify the compounds A, B, C and D. Explain why the milkiness disappears in the last step.

Appearance of milkiness on passing CO₂ in the solution of compound B indicates that compound B is lime water and compound C is CaCO₃. SInce, compound B is obtained by adding H₂O to compound A, therefore, compound A is quicklime, CaO.

The reaction are as follows

(i)

\underset{Calcium}{CaO}

oxide

(A) + H_{2} O \to \underset{Lime water}{Ca {(OH)}_{2}}

(B)

(ii)

\underset{(B)}{Ca {(OH)}_{2}} + {CO}_{2} \to \underset{Calcium carbonate}{{CaCO}_{3}}

(Milkiness) + \underset{(C)}{H_{2} O}

(iii) When excess of CO₂ is passed, milkiness disappears due to the formation of soluble calcium bicarbonate (D).

\underset{Milkiness}{{CaCO}_{3}}

(C) + C O_{2} + H_{2} O \to \underset{Calcium bicarbonate}{Ca {({HCO}_{3})}_{2}}

(Soluble in H_{2} O)

(D)

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