A chemist opened a cupboard and found four bottles containing water solutions, each of which had lost its label.Bottles 1,2,3 contanied colourless sol
A chemist opened a cupboard and found four bottles containing water solutions, each of which had lost its label.Bottles 1,2,3 contanied colourless solution, while bottle 4 contained a blue solution.The labels from the bottles were lying scattered on the floor of the cupboard.They were:
copper (II) sulphate, Hydrochloric acid
lead nitrate , Sodium carbonate
By mixing samples of the contents of the bottles, in pairs , the chemist made the following observations :
Bottle 1 +Bottle 2 `to` White precipitate is formed.
Bottle 1 +Bottle 3 `to` White precipitate is formed.
Bottle 1 +Bottle 4 `to` White precipitate is formed.
Bottle 2 +Bottle 3 `to` Colourless and odourless gas is evolved.
Bottle 2 +Bottle 4 `to` No visible reaction is observed.
Bottle 3 +Bottle 4 `to` Blue precipitate is formed.
With the help of the above observations answer the following questions.
Bottle 3 contains :
A. Copper (II) sulphate
B. Hydrochloric acid
C. Lead nitrate
D. Sodium carbonate
1 Answers
Correct Answer - D
As bottle 2+ bottle 3 gives colourless and odourless gas, it may be carbon dioxide. Generally carbonates are decomposed by acids giving `CO_2` gas.It suggests that bottle 2 and 3 contains sodium carbonate and HCl. Bottle 3+4 gives blue precipitate which confirms the `Cu^(2+)` in either of bottles, `CuSO_4 , CuCl_2 and Cu(NO_3)_2` are soluble and `CuCO_3` is insoluble in water as evident from the reaction.
`Cu^(2+)+CO_3^(2-)toCuCO_3 darr` (blue).Thus blue precipitate must be of copper carbonate.
Hence, bottle 4 is `CuSO_4` , 3 is `Na_2CO_3`, 2 is HCl (from above ) and 1 is `Pb(NO_3)_2` as it gives white precipitate of `PbCl_2` with bottle (2).