Salt Analysis — JEE Main practice

17 questions

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Sample questions with solutions

Q1 · 2026

Consider three metal chlorides $x, y$ and $z$ , where $x$ is water soluble at room temperature, $y$ is sparingly soluble in water at room temperature and $z$ is soluble in hot water. $x, y$ and $z$ are respectively

A.
$AgCl, Hg_2Cl_2$ and $PbCl_2$
B.
$CuCl_2, AgCl$ and $PbCl_2$
C.
$AlCl_3, PbCl_2$ and $BaCl_2$
D.
$MgCl_2, AgCl$ and $AlCl_3$

Answer: B

Why solubility behaviour is the key clue: metal chlorides differ in solubility, so each description ( $x$ , $y$ , $z$ ) must be matched to a chloride known for that exact behaviour.
For $x$ (soluble at room temperature): common chlorides like $MgCl_2$ , $AlCl_3$ , and $CuCl_2$ dissolve readily in cold water, so $x$ should be one of these — here, $CuCl_2$ fits the option set.
For $y$ (only sparingly soluble at room temperature): $AgCl$ and $Hg_2Cl_2$ are the classic examples of chlorides that barely dissolve in cold water, so $y = AgCl$ .
For $z$ (insoluble cold, soluble hot): $PbCl_2$ is the standard example that dissolves poorly in cold water but dissolves noticeably better in hot water — this is exactly why it's used to separate $Pb^{2+}$ from $Ag^+$ and $Hg_2^{2+}$ in group I analysis.
Therefore, matching all three: $x = CuCl_2$ , $y = AgCl$ , $z = PbCl_2$ .

Hence, the answer is B ( $CuCl_2, AgCl$ and $PbCl_2$ ).

Q2 · 2026

In the Group analysis of cations, $\mathrm{Ba}^{2+}$ & $\mathrm{Ca}^{2+}$ are precipitated respectively as

A.
sulphide & sulphide
B.
hydroxide & carbonate
C.
carbonate & carbonate
D.
chromate & sulphide

Answer: C

Why these two ions are grouped together: $\mathrm{Ba}^{2+}$ and $\mathrm{Ca}^{2+}$ both belong to Group V of qualitative cation analysis, and group members are precipitated using the same group reagent.
Since the group V reagent is ammonium carbonate (used along with $NH_4Cl$ and $NH_4OH$ to keep the medium buffered), both ions form insoluble carbonates.

Ba^{2+} + CO_3^{2-} \rightarrow BaCO_3 \downarrow (\text{white})

Ca^{2+} + CO_3^{2-} \rightarrow CaCO_3 \downarrow (\text{white})

Therefore, both cations are precipitated as their respective carbonates, not as sulphides, hydroxides, or chromates (which are tests used for other cations or confirmatory tests, not the group precipitation step).

Hence, the answer is C (carbonate & carbonate).

Q3 · 2026

Given below are two statements:

Statement I: Griss-Ilosvay test is used for the detection of nitrite ion, which involves the use of sulphanilic acid and $\alpha$ -naphthylamine reagent.

Statement II: In the above test, sulphanilic acid is diazotized by the acidified nitrite ion, which on further coupling with $\alpha$ -naphthylamine forms an azo-dye.

In the light of the above statements, choose the correct answer from the options given below

A.
Statement I is true but Statement II is false
B.
Both Statement I and Statement II are true
C.
Both Statement I and Statement II are false
D.
Statement I is false but Statement II is true

Answer: B

Why Statement I is checked first: the Griess–Ilosvay test is the standard qualitative test for detecting nitrite ion ( $NO_2^-$ ), and it is indeed carried out using sulphanilic acid and $\alpha$ -naphthylamine as reagents, so Statement I matches the known procedure.
Why the mechanism in Statement II applies: in acidic medium, the nitrite ion first forms nitrous acid, which reacts with sulphanilic acid in a diazotization step.

HNO_2 + \text{sulphanilic acid} \rightarrow \text{diazonium salt}

Therefore, this diazonium salt undergoes azo coupling with $\alpha$ -naphthylamine to give a coloured azo dye, which is the visible confirmation of nitrite.

\text{diazonium salt} + \alpha\text{-naphthylamine} \rightarrow \text{azo dye (coloured)}

Since both the reagents (Statement I) and the reaction mechanism (Statement II) correctly describe the real test, both statements hold.

Hence, the answer is B (Both Statement I and Statement II are true).

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