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NEET Principles Related to Practical Chemistry MCQs with answers

487 questions on Principles Related to Practical Chemistry for NEET 2027. Below are 15 to try; tap a question to see its answer and explanation.

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15 sample questions

easy

1. An organic compound containing both nitrogen and sulphur gives a blood-red coloration when its sodium fusion extract is treated with FeCl₃. The species responsible for this blood-red color is:

  1. [Fe(CN)₆]³⁻
  2. [Fe(SCN)]²⁺
  3. [Fe(SCN)₆]⁴⁻
  4. [Fe(H₂O)₅(CN)]²⁺
Show answer and explanation

Answer: (B) [Fe(SCN)]²⁺

When both nitrogen and sulphur are present in an organic compound, sodium fusion generates sodium thiocyanate (NaSCN). Upon adding ferric ions (Fe³⁺), thiocyanatoiron(III) complex ions, specifically [Fe(SCN)]²⁺, are formed, which impart a distinct blood-red coloration.

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2. A student dissolves a sample of pure potash alum in water and divides it into two separate test tubes, A and B. To test tube A, aqueous NH₄OH is added dropwise until in excess. To test tube B, aqueous NaOH is added dropwise until in excess. Which of the following correctly describes the final observations in test tubes A and B?

  1. White gelatinous precipitate remains insoluble in A; white gelatinous precipitate dissolves in excess reagent in B
  2. White gelatinous precipitate dissolves in excess reagent in A; white gelatinous precipitate remains insoluble in B
  3. White gelatinous precipitate dissolves in excess reagent in both A and B
  4. White gelatinous precipitate remains insoluble in both A and B
Show answer and explanation

Answer: (A) White gelatinous precipitate remains insoluble in A; white gelatinous precipitate dissolves in excess reagent in B

In both test tubes, Al³⁺ initially precipitates as white gelatinous Al(OH)₃. With NH₄OH (a weak base), Al(OH)₃ does not dissolve even in excess reagent. With NaOH (a strong base), Al(OH)₃ amphoterically dissolves in excess NaOH forming soluble sodium tetrahydroxidoaluminate(III), [Al(OH)₄]⁻ or NaAlO₂.

hard

3. An organic compound 'X' containing C, H, N, and S gave no blood-red coloration with neutral FeCl₃ after Lassaigne's fusion, but gave a violet coloration with sodium nitroprusside and a Prussian blue precipitate with FeSO₄/H₂SO₄. Which experimental condition during the fusion step accounts for this outcome?

  1. Nitric acid was accidentally added before the sodium nitroprusside reagent
  2. The fusion tube was heated insufficiently, preventing the reaction of sulphur
  3. An excess of sodium metal was used during fusion, decomposing NaSCN into NaCN and Na₂S
  4. Excess sulphur in the compound completely consumed all cyanide ions
Show answer and explanation

Answer: (C) An excess of sodium metal was used during fusion, decomposing NaSCN into NaCN and Na₂S

When an organic compound has both N and S, fusion with a limited amount of sodium forms NaSCN (giving a blood-red color with FeCl₃). However, if an excess of sodium is employed, NaSCN is decomposed according to: NaSCN + 2Na → NaCN + Na₂S. The extract contains separate NaCN and Na₂S, thereby testing positive for S (nitroprusside) and N (Prussian blue) separately, without blood-red coloration.

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4. In the layer test (organic solvent test) for the detection of bromine and iodine, the acidified Lassaigne's extract is treated with chlorine water in the presence of CS₂ or CCl₄. Which of the following observations correctly identifies the halogen?

  1. A colorless organic layer confirms Br⁻, while a brown organic layer confirms I⁻
  2. A violet organic layer confirms Br⁻, while an orange organic layer confirms I⁻
  3. An orange/brown organic layer confirms Br⁻, while a violet organic layer confirms I⁻
  4. A deep blue organic layer confirms Br⁻, while a pink organic layer confirms I⁻
Show answer and explanation

Answer: (C) An orange/brown organic layer confirms Br⁻, while a violet organic layer confirms I⁻

Chlorine water oxidises bromide to free bromine (Br₂) and iodide to free iodine (I₂). Bromine dissolves in non-polar organic solvents like CCl₄ or CS₂ to give an orange/brown layer, whereas iodine imparts an intense violet or purple color to the organic layer.

easy

5. What is the role of adding the H₂O₂ solution at the very last moment to the mixture containing KI, dilute H₂SO₄, Na₂S₂O₃, and starch?

  1. To prevent premature reduction of starch indicator by H₂SO₄
  2. To stabilize the thiosulphate against oxidation by atmospheric oxygen
  3. To define an exact zero time (t = 0) since H₂O₂ initiates the oxidation of I⁻
  4. To ensure complete ionization of potassium iodide before reaction
Show answer and explanation

Answer: (C) To define an exact zero time (t = 0) since H₂O₂ initiates the oxidation of I⁻

The redox reaction begins only when both H₂O₂ and I⁻ are brought together in the acidic medium. Adding H₂O₂ last allows the stopwatch to be triggered at the precise moment of mixing (t = 0).

hard

6. 20 mL of 0.1 M oxalic acid is titrated in acid medium and requires 24 mL of KMnO₄ (2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O). The molarity of the KMnO₄ is:

  1. 0.208 M
  2. 0.0333 M
  3. 0.0833 M
  4. 0.0167 M
Show answer and explanation

Answer: (B) 0.0333 M

Moles of MnO₄⁻ = (2/5) × moles of oxalate: M = (2/5) × 0.1 × 20/24 = 0.0333 M.

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7. Which of the following describes the correct crystal system to which Mohr's salt and potash alum belong, respectively?

  1. Monoclinic and Cubic
  2. Triclinic and Orthorhombic
  3. Monoclinic and Hexagonal
  4. Orthorhombic and Cubic
Show answer and explanation

Answer: (A) Monoclinic and Cubic

Mohr's salt, FeSO₄·(NH₄)₂SO₄·6H₂O, crystallizes in the monoclinic crystal system (often described as light green monoclinic prisms). Potash alum, KAl(SO₄)₂·12H₂O, crystallizes in the cubic crystal system, typically forming octahedral crystals.

medium

8. Why is Mohr's salt widely preferred over standard ferrous sulphate heptahydrate (FeSO₄·7H₂O) as a primary standard in redox titrations?

  1. Mohr's salt contains iron in +3 oxidation state which is thermodynamically inert
  2. Mohr's salt does not dissolve completely in aqueous medium
  3. Mohr's salt is significantly more resistant to aerial oxidation in the solid state
  4. Mohr's salt acts as a self-indicator during redox titrations
Show answer and explanation

Answer: (C) Mohr's salt is significantly more resistant to aerial oxidation in the solid state

FeSO₄·7H₂O effloresces and undergoes aerial oxidation to basic ferric sulphate upon exposure to air. In Mohr's salt, the presence of ammonium ions stabilizes the crystal lattice, making it significantly less susceptible to atmospheric oxidation.

easy

9. When hydrated copper(II) sulphate (CuSO₄·5H₂O) dissolves in water, the temperature of the solution decreases. This experimental observation indicates that the enthalpy of solution of hydrated copper sulphate is:

  1. Endothermic with positive ΔH
  2. Exothermic with negative ΔH
  3. Zero because it is a physical process
  4. Exothermic because hydration energy exceeds lattice energy
Show answer and explanation

Answer: (A) Endothermic with positive ΔH

A decrease in temperature during dissolution shows that the system absorbs heat from the surroundings. Hence, the process is endothermic, and the enthalpy of solution (ΔH_sol) has a positive sign.

hard

10. In a coffee-cup calorimeter, 150 mL of 0.8 M HCl is mixed with 150 mL of 0.4 M Ba(OH)₂. If the standard enthalpy of neutralization is -57.3 kJ mol⁻¹ of water formed, what is the total amount of heat evolved in this reaction?

  1. 6.88 kJ
  2. 3.44 kJ
  3. 13.75 kJ
  4. 10.31 kJ
Show answer and explanation

Answer: (A) 6.88 kJ

Moles of H⁺ from HCl = 0.150 L × 0.8 mol L⁻¹ = 0.120 mol. Moles of OH⁻ from Ba(OH)₂ = 2 × (0.150 L × 0.4 mol L⁻¹) = 0.120 mol. Since moles of H⁺ and OH⁻ are equal (0.120 mol), moles of water formed = 0.120 mol. Total heat evolved = 0.120 mol × 57.3 kJ mol⁻¹ = 6.876 kJ ≈ 6.88 kJ.

hard

11. In a study of the kinetics of the reaction between H₂O₂ and I⁻, three experiments yielded the following data: Expt 1: [H₂O₂] = 0.010 M, [I⁻] = 0.010 M, Rate = 1.15 × 10⁻⁶ mol L⁻¹ s⁻¹ Expt 2: [H₂O₂] = 0.020 M, [I⁻] = 0.010 M, Rate = 2.30 × 10⁻⁶ mol L⁻¹ s⁻¹ Expt 3: [H₂O₂] = 0.020 M, [I⁻] = 0.020 M, Rate = 4.60 × 10⁻⁶ mol L⁻¹ s⁻¹ What is the value of the specific reaction rate constant (k)?

  1. 2.30 × 10⁻² L mol⁻¹ s⁻¹
  2. 1.15 × 10⁻² L mol⁻¹ s⁻¹
  3. 1.15 × 10⁻⁴ L mol⁻¹ s⁻¹
  4. 5.75 × 10⁻³ L mol⁻¹ s⁻¹
Show answer and explanation

Answer: (B) 1.15 × 10⁻² L mol⁻¹ s⁻¹

Comparing Expt 1 and 2 shows order with respect to H₂O₂ is 1. Comparing Expt 2 and 3 shows order with respect to I⁻ is 1. Thus, Rate = k[H₂O₂][I⁻]. Using Expt 1: 1.15 × 10⁻⁶ = k(0.010)(0.010) = k(1.0 × 10⁻⁴), which gives k = 1.15 × 10⁻² L mol⁻¹ s⁻¹.

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12. In the standard redox reaction between oxalic acid and potassium permanganate in dilute H₂SO₄, the stoichiometric coefficient of H₂O in the balanced chemical equation is: 2KMnO₄ + 3H₂SO₄ + 5(COOH)₂ → K₂SO₄ + 2MnSO₄ + 10CO₂ + xH₂O The value of x is:

  1. 5
  2. 8
  3. 10
  4. 4
Show answer and explanation

Answer: (B) 8

Balancing hydrogens: from 3H₂SO₄ (6 H) and 5(COOH)₂ (10 H), total H atoms on left = 16. On the right, all H atoms appear as H₂O, so x = 16 / 2 = 8. Oxygen check: left = (2×4) + (3×4) + (5×4) = 8 + 12 + 20 = 40; right = 4 + 8 + 20 + 8 = 40.

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13. 20 mL of 0.1 M oxalic acid (H₂C₂O₄) is exactly neutralised by 30 mL of NaOH solution. The molarity of the NaOH is:

  1. 0.0667 M
  2. 0.15 M
  3. 0.133 M
  4. 0.3 M
Show answer and explanation

Answer: (C) 0.133 M

Oxalic acid is diprotic: 2 × M₁V₁ = M₂V₂, so M₂ = 2 × 0.1 × 20/30 = 0.133 M.

hard

14. Equal volumes of 0.1 M aqueous solutions of Mohr's salt and potash alum are prepared. What is the theoretical ratio of the total molar concentration of sulphate ions (SO₄²⁻) in the Mohr's salt solution to that in the potash alum solution? (Assume potash alum is formulated as KAl(SO₄)₂·12H₂O)

  1. 3 : 2
  2. 2 : 1
  3. 1 : 2
  4. 1 : 1
Show answer and explanation

Answer: (D) 1 : 1

One formula unit of Mohr's salt, FeSO₄·(NH₄)₂SO₄·6H₂O, releases 2 SO₄²⁻ ions upon dissociation, so a 0.1 M solution has [SO₄²⁻] = 0.2 M. One formula unit of potash alum, KAl(SO₄)₂·12H₂O, also releases 2 SO₄²⁻ ions, so its 0.1 M solution has [SO₄²⁻] = 0.2 M. The ratio is 0.2 : 0.2 = 1 : 1.

hard

15. A 0.50 g impure sample of oxalic acid dihydrate, (COOH)₂·2H₂O (molar mass = 126 g mol⁻¹), was dissolved in dilute H₂SO₄ and titrated against 0.02 M KMnO₄ solution. The endpoint was reached at 30.0 mL of KMnO₄. What is the percentage purity of the oxalic acid sample?

  1. 75.6%
  2. 37.8%
  3. 94.5%
  4. 18.9%
Show answer and explanation

Answer: (B) 37.8%

Equivalents of KMnO₄ = M × n-factor × V(L) = 0.02 × 5 × 0.030 = 0.003 equiv. Equivalents of oxalic acid = 0.003. For (COOH)₂·2H₂O, n-factor = 2, so moles = 0.003 / 2 = 0.0015 mol. Mass of pure dihydrate = 0.0015 × 126 = 0.189 g. Percentage purity = (0.189 / 0.50) × 100% = 37.8%.

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