JEE Chemistry Salt Analysis MCQs Set 06

Practice MCQs for JEE Chemistry Salt Analysis

Access targeted multiple-choice questions for Salt Analysis designed to align with the latest JEE academic syllabus for JEE Chemistry. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

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Multiple Choice Questions

Question 1: The Lassaigne's extract is boiled with dil. HNO3 before testing for halogens because:
(a) silver halides are soluble in HNO3
(b) Na2S and NaCN are decomposed by HNO3
(c) Ag2S is soluble in HNO3
(d) AgCN is soluble in HNO3
Show Answer & Explanation

Answer: (b) Na2S and NaCN are decomposed by HNO3

Explanation:
1. Na2S and NaCN would give black Ag2S and white AgCN with AgNO3, hiding the silver halide precipitate.
2. Boiling with HNO3 decomposes them to H2S and HCN, which escape.
3. Answer: (b)

Teacher's Note:
1. Boil to remove interfering S2− and CN−.
2. Do this in a fume cupboard.

Question 2: For preparing 250 mL of N/20 solution of Mohr's salt, the amount of Mohr's salt needed is:
(a) 9.8 g
(b) 4.9 g
(c) 19.6 g
(d) 3.2 g
Show Answer & Explanation

Answer: (b) 4.9 g

Explanation:
1. Equivalent mass of Mohr's salt FeSO4(NH4)2SO4·6H2O = 392.
2. Mass = N × E × V(L) = (1/20) × 392 × 0.25 = 4.9 g.
3. Answer: (b)

Teacher's Note:
1. Mass = normality × equivalent mass × volume in litres.
2. Mohr's salt is a primary-standard-like reducing agent.

Question 3: Potassium chromate solution is added to an aqueous solution of a metal chloride. The precipitate thus obtained is insoluble in acetic acid. When the precipitate is subjected to flame test the colour of the flame is:
(a) lilac
(b) apple green
(c) crimson red
(d) golden yellow
Show Answer & Explanation

Answer: (b) apple green

Explanation:
1. BaCrO4 is a yellow precipitate that is insoluble in acetic acid, so the cation is Ba2+.
2. Ba2+ gives an apple green flame.
3. Answer: (b)

Teacher's Note:
1. PbCrO4 is also yellow but dissolves in NaOH.
2. Sr gives crimson, K lilac and Na golden yellow.

Question 4: Three separate samples of a solution of a single salt gave these results. One formed a white precipitate with excess ammonia solution, one formed a white precipitate with dil. NaCl solution and one formed a black precipitate with H2S. The salt could be:
(a) AgNO3
(b) Pb(NO3)2
(c) Hg(NO3)2
(d) MnSO4
Show Answer & Explanation

Answer: (b) Pb(NO3)2

Explanation:
1. Pb(NO3)2 + NH4OH gives white Pb(OH)2 insoluble in excess.
2. Pb2+ + NaCl gives white PbCl2.
3. Pb2+ + H2S gives black PbS.
4. AgNO3 gives a clear solution with excess ammonia, Hg(NO3)2 gives no chloride precipitate, and MnSO4 gives no black sulphide in these tests.
5. Answer: (b)

Teacher's Note:
1. Pb2+ behaves this way in all three tests.
2. Check each option against all three clues.

Question 5: The salt used for performing the 'bead' test in qualitative inorganic analysis is:
(a) K2SO4·Al2(SO4)3·24H2O
(b) FeSO4·(NH4)2SO4·4H2O
(c) Na(NH4)HPO4·4H2O
(d) CaSO4·2H2O
Show Answer & Explanation

Answer: (c) Na(NH4)HPO4·4H2O

Explanation:
1. Microcosmic salt, Na(NH4)HPO4·4H2O, on heating gives NaPO3 (a glassy bead).
2. The metaphosphate bead combines with metal oxides to give coloured orthophosphates, for example blue NaCoPO4.
3. Answer: (c)

Teacher's Note:
1. Borax bead and microcosmic salt bead are the two bead tests.
2. The colour of the bead identifies the metal.

Question 6: A gas 'X' is passed through water to form a saturated solution. The aqueous solution on treatment with AgNO3 gives a white precipitate. The saturated aqueous solution also dissolves magnesium ribbon with evolution of a colourless gas 'Y'. Identify X and Y.
(a) X = CO2, Y = Cl2
(b) X = Cl2, Y = CO2
(c) X = Cl2, Y = H2
(d) X = H2, Y = Cl2
Show Answer & Explanation

Answer: (c) X = Cl2, Y = H2

Explanation:
1. Cl2 + H2O → HCl + HOCl; HCl gives white AgCl with AgNO3.
2. Mg + 2HCl → MgCl2 + H2, so Y is hydrogen.
3. Answer: (c)

Teacher's Note:
1. Chlorine water is acidic.
2. Active metals liberate H2 from acids.

Question 7: In Kjeldahl's method for the estimation of N2, potassium sulphate and copper sulphate are used. On the basis of their functions which of the following statements is/are correct? I. Potassium sulphate raises the boiling point and ensures complete reaction. II. Copper sulphate acts as catalyst. III. Potassium sulphate acts as catalyst and copper sulphate raises the boiling point.
(a) Only III is correct
(b) I and II are correct
(c) Only II is correct
(d) None is correct
Show Answer & Explanation

Answer: (b) I and II are correct

Explanation:
1. K2SO4 raises the boiling point of the digestion mixture so the reaction is complete.
2. CuSO4 acts as the catalyst.
3. So statements I and II are correct and III is wrong.
4. Answer: (b)

Teacher's Note:
1. Know the role of each reagent in digestion.
2. Statement III reverses the roles.

Question 8: Aniline is usually purified by:
(a) chromatographic technique
(b) steam distillation
(c) addition of oxalic acid
(d) fractional crystallisation
Show Answer & Explanation

Answer: (b) steam distillation

Explanation:
1. Aniline is immiscible with water, volatile in steam and decomposes near its boiling point.
2. Steam distillation lets it distil below 100 °C without decomposition.
3. Answer: (b)

Teacher's Note:
1. Steam distillation suits liquids immiscible with water.
2. It is used for aniline, nitrobenzene and essential oils.

Question 9: Prussian blue is formed when:
(a) ferrous sulphate reacts with FeCl3
(b) ferric sulphate reacts with K4Fe(CN)6
(c) ferrous ammonium sulphate reacts with FeCl3
(d) ammonium sulphate reacts with NO2
Show Answer & Explanation

Answer: (b) ferric sulphate reacts with K4Fe(CN)6

Explanation:
1. 2Fe2(SO4)3 + 3K4[Fe(CN)6] → Fe4[Fe(CN)6]3 + 6K2SO4.
2. The product, ferric ferrocyanide, is Prussian blue.
3. Answer: (b)

Teacher's Note:
1. Fe3+ + ferrocyanide gives Prussian blue.
2. Same compound forms in the Lassaigne nitrogen test.

Question 10: An organic compound has C and H percentage in the ratio 6 : 1 and C and O percentage in the ratio 3 : 4. The compound is:
(a) HCHO
(b) CH3OH
(c) CH3CH2OH
(d) (COOH)2
Show Answer & Explanation

Answer: (a) HCHO

Explanation:
1. Take C : H : O by mass = 6 : 1 : 8.
2. Moles: C = 6/12 = 0.5, H = 1/1 = 1, O = 8/16 = 0.5.
3. Ratio C : H : O = 1 : 2 : 1, so the empirical formula is CH2O, matching HCHO.
4. Answer: (a)

Teacher's Note:
1. Divide mass ratios by atomic masses.
2. CH2O is also the empirical formula of glucose.

Question 11: The presence of magnesium is confirmed in qualitative analysis by the formation of a white crystalline precipitate of:
(a) Mg(HCO3)2
(b) MgNH4PO4
(c) MgNH4(HCO3)3
(d) MgCO3
Show Answer & Explanation

Answer: (b) MgNH4PO4

Explanation:
1. MgCl2 + Na2HPO4 + NH4OH → Mg(NH4)PO4 + 2NaCl + H2O.
2. The white crystalline precipitate is magnesium ammonium phosphate.
3. Answer: (b)

Teacher's Note:
1. Group VI test for Mg2+.
2. Ammonium chloride prevents Mg(OH)2 formation.

Question 12: A salt which gives CO2 with hot conc. H2SO4 and also decolourises acidified KMnO4 on warming is:
(a) HCO3−
(b) CO32−
(c) oxalate
(d) acetate
Show Answer & Explanation

Answer: (c) oxalate

Explanation:
1. Oxalate with hot conc. H2SO4 gives CO + CO2 + H2O.
2. Oxalate reduces warm acidified KMnO4 (purple to colourless) and itself gives CO2.
3. Carbonate gives CO2 but does not reduce permanganate.
4. Answer: (c)

Teacher's Note:
1. Oxalate is a reducing anion.
2. Carbonate and bicarbonate give CO2 with dilute acids.

Question 13: [X] + H2SO4 gives [Y], a colourless gas with irritating smell. [Y] + K2Cr2O7 + H2SO4 gives a green solution. [X] and [Y] are:
(a) SO32−, SO2
(b) Cl−, HCl
(c) S2−, H2S
(d) CO32−, CO2
Show Answer & Explanation

Answer: (a) SO32−, SO2

Explanation:
1. Sulphite + dilute H2SO4 gives SO2, a colourless gas with a suffocating smell.
2. SO2 reduces orange Cr2O72− to green Cr3+.
3. H2S has a rotten egg smell and CO2 is odourless; HCl does not reduce dichromate.
4. Answer: (a)

Teacher's Note:
1. SO2 turns acidified dichromate green.
2. H2S also turns it green, but it smells of rotten eggs.

Question 14: The sodium extract prepared from sulphanilic acid contains SCN−. It gives blood red colouration with:
(a) a mixture of Na2S and CS2
(b) FeCl3
(c) FeSO4
(d) Na2SO3
Show Answer & Explanation

Answer: (b) FeCl3

Explanation:
1. Sulphanilic acid contains N and S, so fusion with sodium gives NaSCN.
2. SCN− reacts with Fe3+ (FeCl3) to give blood red [Fe(SCN)]2+.
3. Fe2+ does not give this colour.
4. Answer: (b)

Teacher's Note:
1. When both N and S are present, SCN− forms.
2. Use ferric salt, not ferrous.

Question 15: An orange precipitate of group II is dissolved in conc. HCl; the solution when treated with excess of water turns milky due to formation of:
(a) Sn(OH)Cl
(b) Sb(OH)Cl2
(c) SbOCl
(d) Sb(OH)2Cl
Show Answer & Explanation

Answer: (c) SbOCl

Explanation:
1. The orange sulphide is Sb2S3, which dissolves in conc. HCl as SbCl3.
2. On dilution SbCl3 hydrolyses: SbCl3 + H2O → SbOCl + 2HCl, giving a white milky precipitate.
3. Answer: (c)

Teacher's Note:
1. SbCl3 and BiCl3 hydrolyse to oxychlorides.
2. Sb2S3 is orange, SnS is brown.

Question 16: A substance on treatment with dil. H2SO4 liberates a colourless gas which produces (I) turbidity with baryta water and (II) turns acidified dichromate solution green. The reaction indicates the presence of:
(a) CO32−
(b) S2−
(c) SO32−
(d) NO3−
Show Answer & Explanation

Answer: (c) SO32−

Explanation:
1. SO2 turns baryta water milky (BaSO3) and reduces orange dichromate to green Cr3+.
2. CO2 turns baryta water milky but does not affect dichromate, so carbonate is ruled out.
3. Answer: (c)

Teacher's Note:
1. Both tests together identify SO2.
2. Sulphite is the source.

Question 17: Which of the following gives a precipitate with Pb(NO3)2 but not with Ba(NO3)2?
(a) Sodium chloride
(b) Sodium acetate
(c) Sodium nitrate
(d) Sodium hydrogen phosphate
Show Answer & Explanation

Answer: (a) Sodium chloride

Explanation:
1. NaCl + Pb(NO3)2 gives white PbCl2, while BaCl2 is soluble.
2. Sodium hydrogen phosphate would precipitate both, and acetate and nitrate salts of Pb and Ba are soluble.
3. Answer: (a)

Teacher's Note:
1. PbCl2 is sparingly soluble in cold water.
2. Barium chloride is highly soluble.

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Multiple Choice Questions (MCQs) for JEE Chemistry Salt Analysis

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