The correct answer is 40
Mass of organic compound = 0.25 g
Mass of AgCl = 0.40 g
%\(Cl=\frac{35.5×mass\ of AgCl}{143.5×(mass\ of\ organic\ compound)}×100\)
\(=\frac{35.5×0.40×100}{143.5×0.25 }\)
= 39.581 ≃ 40%
Therefore , Cl = 40 %
| List-I (Name of the test) | List-II (Reaction sequence involved) [M is metal] | |
|---|---|---|
| (A) Borax bead test | I | ![]() |
| (B) Charcoal cavity test | II | ![]() |
| (C) Cobalt nitrate test | III | ![]() |
| (D) Flame test | IV | ![]() |
Given below are two statements:
Statement (I):
are isomeric compounds.
Statement (II):
are functional group isomers.
In the light of the above statements, choose the correct answer from the options given below:
Among the following cations, the number of cations which will give characteristic precipitate in their identification tests with
\(K_4\)[Fe(CN)\(_6\)] is : \[ {Cu}^{2+}, \, {Fe}^{3+}, \, {Ba}^{2+}, \, {Ca}^{2+}, \, {NH}_4^+, \, {Mg}^{2+}, \, {Zn}^{2+} \]