JEE Chemistry Analytical Chemistry MCQs Set 01

Answer: (c) m-cresol

Explanation:
1. Solubility in NaOH but not in NaHCO₃ shows a weakly acidic group such as phenolic OH.
2. m-Cresol forms sodium 3-methylphenoxide with NaOH; oleic acid would also dissolve in NaHCO₃, o-toluidine would dissolve in HCl and benzamide in none.

Teacher's Note:
1. Phenols dissolve in NaOH but not in NaHCO₃; carboxylic acids dissolve in both.

Answer: (b) phenol

Explanation:
1. Insolubility in dil. HCl rules out the amines.
2. Solubility in NaOH and decolourisation of bromine water (forming 2,4,6-tribromophenol) show a phenol.

Teacher's Note:
1. Phenol is acidic and is brominated readily by bromine water.

Answer: (d) (CH₃)₃C–Cl

Explanation:
1. A precipitate of AgCl forms only if the halide ionises easily to a stable carbocation.
2. (CH₃)₃C–Cl gives the stable tertiary carbocation, whereas vinyl chloride, CCl₄ and CHCl₃ do not ionise readily.

Teacher's Note:
1. Tertiary alkyl halides react with AgNO₃ readily through SN1.

Answer: (c) A-Q, B-P, C-S, D-R

Explanation:
1. Lysine (an amino acid) gives a violet colour with ninhydrin.
2. Furfural with 1-naphthol and H₂SO₄ (Molisch test) gives a violet colour, benzyl alcohol is oxidised by ceric ammonium nitrate, and styrene decolourises KMnO₄ (Baeyer test).

Teacher's Note:
1. Link each functional group with its characteristic test.

Answer: (b) 25 mL

Explanation:
1. 2HCl + Na₂CO₃ → 2NaCl + H₂O + CO₂: molarity of HCl = (30 × 0.1 × 2)/25 = 0.24 M.
2. Moles of NaOH = 30 × 0.2 = 6 mmol, so the volume of HCl = 6/0.24 = 25 mL.

Teacher's Note:
1. Equate milliequivalents of acid and base.

Answer: (a) 4 g

Explanation:
1. Milliequivalents: 50 × 0.5 × 2 = 25 × M(NaOH), so M(NaOH) = 2 M.
2. In 50 mL there are 2 × 0.05 = 0.1 mol, that is 0.1 × 40 = 4 g of NaOH.

Teacher's Note:
1. Oxalic acid is dibasic, so its normality is twice its molarity.

Answer: (b) 10%

Explanation:
1. H₂SO₄ taken = 60 × 0.1 = 6 mmol; NaOH used = 2 mmol, which neutralises 1 mmol of H₂SO₄.
2. H₂SO₄ that reacted with NH₃ = 5 mmol, so NH₃ = 10 mmol = N = 10 mmol, mass 0.14 g.
3. Percentage of N = 0.14/1.4 × 100 = 10%.

Teacher's Note:
1. Each mole of H₂SO₄ neutralised by ammonia corresponds to 2 mol of nitrogen.

Answer: (c) 23.7

Explanation:
1. HCl taken = 2 mmol and NaOH used = 1.5 mmol, so HCl used by NH₃ = 0.5 mmol.
2. Nitrogen = 0.5 mmol × 14 = 7 mg, so the percentage is 7/29.5 × 100 = 23.7.

Teacher's Note:
1. Back titration gives the acid consumed by ammonia.

Answer: (c) One

Explanation:
1. EDTA is a hexadentate ligand, with two N and four carboxylate O donor atoms.
2. One EDTA molecule therefore occupies all six octahedral positions.

Teacher's Note:
1. Number of ligands = coordination number/denticity.

Answer: (a) Chlorophyll is a green pigment in plants and contains calcium

Explanation:
1. Chlorophyll is a magnesium complex of a porphyrin, not a calcium complex.
2. The other three statements are correct: Fe in haemoglobin, Co in vitamin B₁₂ and Zn in carboxypeptidase-A.

Teacher's Note:
1. Chlorophyll has Mg, haemoglobin has Fe, vitamin B₁₂ has Co.

Answer: (a) Fe₄[Fe(CN)₆]₃

Explanation:
1. Nitrogen is converted to NaCN on fusion with sodium, and NaCN gives Na₄[Fe(CN)₆] with FeSO₄.
2. With FeCl₃ this gives the Prussian blue precipitate, Fe₄[Fe(CN)₆]₃.

Teacher's Note:
1. Prussian blue is the positive Lassaigne test for nitrogen.

Answer: (b) addition of NH₄OH in the presence of NH₄Cl when both Cr(OH)₃ and Fe(OH)₃ are precipitated and on adding Br₂ water and NaOH, Cr(OH)₃ dissolves

Explanation:
1. Both hydroxides are precipitated by NH₄OH in the presence of NH₄Cl.
2. Treating the precipitate with NaOH and bromine water oxidises Cr(OH)₃ to soluble yellow chromate, while Fe(OH)₃ remains insoluble; conc. HCl dissolves both hydroxides, so (b) alone is correct.

Teacher's Note:
1. Chromium(III) is oxidised to yellow chromate in alkaline bromine water.