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JEE • CHEMISTRY • PRACTICAL CHEMISTRY

Practical chemistry practice · Medium

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 61-84 of 84 questions
61 Medium
61 of 84
Analyzing +4 | -1
An aqueous solution of colourless metal sulphate M,M, gives a white ppt. with NH4OH.\mathrm{NH}_4 \mathrm{OH}. This was soluble in excess of NH4OH.\mathrm{NH}_4 \mathrm{OH}. On passing H2S\mathrm{H}_2 \mathrm{S} through this solution a white ppt is formed. The metal MM in the salt is
View answer and explanation
Correct option: D
Zn\mathrm{Zn}
Option D (Zn) is correct because Zinc sulfate (ZnSO4) forms a white precipitate with NH4OH that dissolves in excess, and then forms a white precipitate of ZnS with H2S.
62 Medium
62 of 84
Understanding +4 | -1
Which method(s) can be used to determine the concentration of HNO3\mathrm{HNO}_3 in an aqueous solution of nitric acid?

I) titration with a standard base

II) titration with a standard oxidizing agent

III) precipitation with Ag+\mathrm{A g}^+
View answer and explanation
Correct option: A
I only
Option A (I only) is correct because the concentration of HNO3, being an acid, can be determined by titration with a standard base.
63 Medium
63 of 84
Remembering +4 | -1
......... is formed when potassium iodide is heated with conc. H2SO4\mathrm{H}_2 \mathrm{SO}_4
View answer and explanation
Correct option: B
I2I_2
Option B (I2I_2) is correct because when potassium iodide is heated with concentrated H2SO4\mathrm{H}_2 \mathrm{SO}_4, iodine (I2I_2) is produced.
64 Medium
64 of 84
Analyzing +4 | -1
An aqueous solution of FeSO4⋅Al2(SO4)3\mathrm{FeSO}_4 \cdot \mathrm{Al}_2\left(\mathrm{SO}_4\right)_3 and chrome alum is heated with excess of Na2O2\mathrm{Na}_2 \mathrm{O}_2 and filtered the materials obtained are
View answer and explanation
Correct option: C
A yellow filtrate and a brown residue
The correct answer is C (A yellow filtrate and a brown residue) because the yellow filtrate contains sodium chromate (Na₂CrO₄) formed by oxidation of Cr³⁺, and the brown residue is due to the formation of basic ferric sulfate from the oxidation of FeSO₄.
65 Medium
65 of 84
Remembering +4 | -1
Which of the following does not give NH3{NH}_3 on heating with NaOHNaOH solution?
View answer and explanation
Correct option: B
Aniline
Option B (Aniline) is correct because aniline does not have an amide group that can be hydrolyzed by NaOH to release ammonia.
66 Medium
66 of 84
Analyzing +4 | -1
When a substance AA reacts with water it produces a combustible gas BB and a solution of substance CC in water. When another substance DD reacts with this solution of C,C, it also produces the same gas BB on warming but DD can produce gas BB on reaction with dilute sulphuric acid at room temperature. AA imparts a deep golden yellow colour a smokeless flame to Bunsen burner. A,B,CA, B, C and DD respectively are:
View answer and explanation
Correct option: A
Na,H2,NaOH,Zn\mathrm{Na}, \mathrm{H}_2, \mathrm{NaOH}, \mathrm{Zn}
The correct answer is A (Na,H2,NaOH,Zn\mathrm{Na}, \mathrm{H}_2, \mathrm{NaOH}, \mathrm{Zn}) because it correctly identifies the substances based on their reactions with water and acids, and the flame test.
67 Medium
67 of 84
Remembering +4 | -1
The compound formed in the borax bead test of Cu2+\mathrm{Cu}^{2+} ion in oxidising flame is:
View answer and explanation
Correct option: C
Cu(BO2)2\mathrm{Cu}\left(\mathrm{BO}_2\right)_2
Option C (Cu(BO2)2\mathrm{Cu}(\mathrm{BO}_2)_2) is correct because in the borax bead test, Cu2+\mathrm{Cu}^{2+} ions react with borax to form cupric metaborate, which has a characteristic color.
68 Medium
68 of 84
Remembering +4 | -1
Phenolphltalein indicator is prepared by
dissolving phenolpthalein in
View answer and explanation
Correct option: B
Ethyl alcohol
Option B (Ethyl alcohol) is correct because phenolphthalein is soluble in ethyl alcohol and is typically dissolved in it to prepare the indicator solution.
69 Medium
69 of 84
Remembering +4 | -1
Which one of the following salt give green coloured flame when the salt is tested by Pt wire
View answer and explanation
Correct option: A
Barium salt
Option A (Barium salt) is correct because barium salts produce a green-colored flame in the flame test due to the excitation of barium ions.
70 Medium
70 of 84
Remembering +4 | -1
On passing a gas, 'XX', through Nessler's regent, a brown precipitate is obtained. The gas 'XX' is
View answer and explanation
Correct option: B
NH3{NH}_3
Option B (NH3{NH}_3) is correct because ammonia reacts with Nessler's reagent to form a brown precipitate.
71 Medium
71 of 84
Remembering +4 | -1
The presence of NH4+\mathrm{NH}_4^{+} radical in solution can be detected by
View answer and explanation
Correct option: D
Nessler's reagent
Option D (Nessler's reagent) is correct because Nessler's reagent (K₂HgI₄) reacts with ammonium ions (NH₄⁺) to form a brown precipitate, indicating the presence of ammonium ions.
72 Medium
72 of 84
Application +4 | -1
In Carius method of estimation of halogens, 0.5 g of an organic compound gave 0.376 g of AgBr. The percentage of bromine in the compound is (Atomic mass of Ag = 108 u and Br = 80 u)
View answer and explanation
Correct option: D
0.32
Option D (0.32) is correct because the percentage of bromine is calculated using the ratio of the mass of bromine to the mass of AgBr, multiplied by the ratio of the mass of AgBr to the mass of the organic compound, and then by 100.
73 Medium
73 of 84
Remembering +4 | -1
The metal ion which is precipitated when H2S\mathrm{H}_2 \mathrm{S} is passed with HCl.\mathrm{HCl}.
View answer and explanation
Correct option: C
Cd2+\mathrm{Cd}^{2+}
Option C (Cd2+\mathrm{Cd}^{2+}) is correct because Cadmium ions are precipitated as sulfides when H2S is passed through an acidic solution (HCl).
74 Medium
74 of 84
Analyzing +4 | -1
Which of the following pairs of cations can not be separated by using NH3\mathrm{NH}_3 solution.
View answer and explanation
Correct option: B
Zn+2,Cu+2\mathrm{Zn}^{+2}, \mathrm{Cu}^{+2}
Option D (Al+3,As3+\mathrm{Al}^{+3}, \mathrm{As}^{3+}) is correct because both aluminum and arsenic ions form precipitates with ammonia solution.
75 Medium
75 of 84
Remembering +4 | -1
In borax bead test, which of the following compound is formed
View answer and explanation
Correct option: A
Meta borate
Option A (Meta borate) is correct because in the borax bead test, metal oxides react with borax to form metaborates, which impart characteristic colors.
76 Medium
76 of 84
Understanding +4 | -1
Which of the following pairs of ions would be expected to form precipitate when dilute solution are mixed?
View answer and explanation
Correct option: D
Fe3+,PO43−\mathrm{Fe}^{3+}, \mathrm{PO}_4^{3-}
Option D (Fe3+, PO43-) is correct because Fe3+ and PO43- ions react to form insoluble iron(III) phosphate (FePO4), which precipitates out of the solution.
77 Medium
77 of 84
Remembering +4 | -1
When acetic acid and potassium ferrocyanide are added to copper salt, the colour of the precipitate is
View answer and explanation
Correct option: C
Brown
Option C (Brown) is correct because the reaction between copper salt and potassium ferrocyanide results in a reddish brown precipitate of copper ferrocyanide.
78 Medium
78 of 84
Remembering +4 | -1
MnO2MnO_2 and H2SO4H_2SO_4 added to NaCl, the greenish yellow gas liberated is
View answer and explanation
Correct option: A
Cl2Cl_2
Option A (Cl2Cl_2) is correct because the reaction of MnO2MnO_2 and H2SO4H_2SO_4 with NaCl produces chlorine gas, which is greenish-yellow.
79 Medium
79 of 84
Understanding +4 | -1
Sometimes yellow turbidity appears while passing H2S\mathrm{H}_2 \mathrm{S} gas even in the absence of II group radicals. This is because of
View answer and explanation
Correct option: C
The oxidation of H2S\mathrm{H}_2 \mathrm{S} gas by some acid radicals
Option C (The oxidation of H2S\mathrm{H}_2 \mathrm{S} gas by some acid radicals) is correct because the yellow turbidity is due to the formation of elemental sulfur by the oxidation of H2SH_2S by oxidizing agents present.
80 Medium
80 of 84
Remembering +4 | -1
An orange precipitate in the second group is given by
View answer and explanation
Correct option: B
Sb3+\mathrm{Sb}^{3+}
Option B (Sb3+\mathrm{Sb}^{3+}) is correct because antimony(III) sulfide (Sb2S3\mathrm{Sb}_2 \mathrm{S}_3) precipitates as an orange solid in the second group analysis.
81 Medium
81 of 84
Remembering +4 | -1
Mark the compound which is soluble in hot water.
View answer and explanation
Correct option: A
Lead chloride
Option A (Lead chloride) is correct because lead chloride (PbCl2PbCl_2) is known to be soluble in hot water due to increased kinetic energy breaking the lattice structure.
82 Medium
82 of 84
Remembering +4 | -1
NiS\text{NiS} is separated from ZnS\mathrm{ZnS} by treating with
View answer and explanation
Correct option: B
conc. HCl\mathrm{HCl}
Option B (conc. HCl\mathrm{HCl}) is correct because NiS\mathrm{NiS} is insoluble in conc. HCl\mathrm{HCl} while ZnS\mathrm{ZnS} is soluble.
83 Medium
83 of 84
Remembering +4 | -1
When H2S\mathrm{H}_2 \mathrm{S} gas is passed in a metal sulphate solution in presence of NH4OH,\mathrm{NH}_4 \mathrm{OH}, a white precipitate is produced. The metal is identified as
View answer and explanation
Correct option: A
Zn\mathrm{Zn}
Option A is correct because Zinc (Zn) ions react with H2S in the presence of NH4OH to form a white precipitate of Zinc sulfide (ZnS).
84 Medium
84 of 84
Remembering +4 | -1
Which of the following statements is correct about phenolphthalein:

(A) It is a phthalein dye.

(B) It is used as an acid base indicator

(C) It is obtained by condensation of phenol with phthalic anhydride in presence of conc. H2SO4\mathrm{H}_2 \mathrm{SO}_4
View answer and explanation
Correct option: D
A, B, C
Option D (A, B, C) is correct because all three statements about phenolphthalein are accurate and well-established facts.
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