CBSE Class 12 Chemistry Board Paper 1 — MCQs with Answers

8 solved questions from this paper

Answer and worked explanation shown for each. The remaining 12 are timed and scored when you sign in free.

1. Q1Easy

   18 g of glucose (molar mass 180 g mol⁻¹) is dissolved in 500 g of water. The molality of the solution is:

   A.0.10 mol kg⁻¹

   B.2.00 mol kg⁻¹

   C.0.20 mol kg⁻¹✓ Correct

   D.0.05 mol kg⁻¹

   Answer: C. 0.20 mol kg⁻¹

   Explanation: Molality = moles of solute ÷ mass of SOLVENT in kilograms. Moles of glucose = 18/180 = 0.10 mol and the solvent mass is 0.500 kg, so m = 0.10/0.500 = 0.20 mol kg⁻¹. The common slip is to divide 0.10 mol by 1 kg (or by the 500 as if it were already kilograms), which gives the too-small value 0.10 mol kg⁻¹.

2. Q2Easy

   In the reaction 2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O, the number of electrons gained by each permanganate ion is:

   A.2

   B.5✓ Correct

   C.3

   D.10

   Answer: B. 5

   Explanation: Manganese goes from +7 in MnO₄⁻ to +2 in Mn²⁺, a fall of 5 units, so each permanganate ion accepts 5 electrons. The value 10 is the TOTAL number of electrons transferred by the two permanganate ions in the balanced equation, not the number gained per ion; 3 corresponds to the +7 → +4 change that occurs in neutral or faintly alkaline medium instead.

3. Q3Easy

   Which of the following aldehydes undergoes the Cannizzaro reaction on warming with concentrated NaOH?

   A.ethanal

   B.propanal

   C.benzaldehyde✓ Correct

   D.2-methylpropanal

   Answer: C. benzaldehyde

   Explanation: Cannizzaro disproportionation requires an aldehyde with NO α-hydrogen, so the hydroxide adds to the carbonyl instead of removing a proton. Benzaldehyde's carbonyl carbon is attached to a ring carbon and a hydrogen, leaving no α-hydrogen at all. The other three each carry α-hydrogens and therefore give aldol condensation products with base rather than the acid-plus-alcohol pair.

4. Q4Medium

   For the cell Zn(s) | Zn²⁺(0.1 M) || Cu²⁺(1.0 M) | Cu(s), E°cell = 1.10 V at 298 K. The EMF of the cell is:

   A.1.13 V✓ Correct

   B.1.07 V

   C.1.16 V

   D.1.04 V

   Answer: A. 1.13 V

   Explanation: The Nernst equation for this two-electron cell is Ecell = E°cell − (0.059/2) log([Zn²⁺]/[Cu²⁺]). Substituting, log(0.1/1.0) = −1, so Ecell = 1.10 − 0.0295(−1) = 1.13 V. Writing the quotient upside down as [Cu²⁺]/[Zn²⁺] flips the sign and produces the trap value 1.07 V; diluting the product ion must raise the EMF, not lower it.

5. Q5Medium

   The rate of a first-order reaction is 0.040 mol L⁻¹ s⁻¹ when the concentration of the reactant is 0.20 mol L⁻¹. The half-life of the reaction is:

   A.17.3 s

   B.0.139 s

   C.5.00 s

   D.3.47 s✓ Correct

   Answer: D. 3.47 s

   Explanation: For a first-order reaction rate = k[A], so k = 0.040/0.20 = 0.20 s⁻¹. Then t½ = 0.693/k = 0.693/0.20 = 3.47 s. Multiplying 0.693 by k instead of dividing gives 0.139 s, and using 1/k alone gives 5.00 s — the 0.693 factor is what converts the time constant into a half-life.

6. Q6Medium

   The IUPAC name of the complex [Pt(NH₃)₂Cl₂] is:

   A.dichloridodiammineplatinum(II)

   B.diamminedichloridoplatinum(IV)

   C.diamminedichloridoplatinate(II)

   D.diamminedichloridoplatinum(II)✓ Correct

   Answer: D. diamminedichloridoplatinum(II)

   Explanation: Ligands are cited in alphabetical order of their names ignoring the multiplying prefix, so ammine comes before chlorido. Two neutral NH₃ and two Cl⁻ on an overall neutral complex fix platinum at +2, giving diamminedichloridoplatinum(II). The ending -platinate is reserved for complexes that are anions, and this complex carries no charge at all.

7. Q7Hard

   The vapour pressure of pure benzene at a certain temperature is 640 mm Hg. On dissolving 2.5 g of a non-volatile, non-electrolyte solute in 39 g of benzene (molar mass 78 g mol⁻¹), the vapour pressure falls to 600 mm Hg. The molar mass of the solute is:

   A.75 g mol⁻¹✓ Correct

   B.80 g mol⁻¹

   C.64 g mol⁻¹

   D.150 g mol⁻¹

   Answer: A. 75 g mol⁻¹

   Explanation: Raoult's law gives the relative lowering of vapour pressure as the mole fraction of solute: (640 − 600)/640 = 0.0625. With benzene = 39/78 = 0.50 mol, solving n/(n + 0.50) = 0.0625 gives n = 0.03333 mol, so M = 2.5/0.03333 = 75 g mol⁻¹. Using the dilute approximation n ≈ 0.0625 × 0.50 = 0.03125 gives the slightly high value 80 g mol⁻¹, which is not accurate enough at a 6% mole fraction.

8. Q8Hard

   The rate constant of a reaction doubles when the temperature is raised from 300 K to 310 K. Taking R = 8.314 J K⁻¹ mol⁻¹, the activation energy is closest to:

   A.26.8 kJ mol⁻¹

   B.107.2 kJ mol⁻¹

   C.53.6 kJ mol⁻¹✓ Correct

   D.5.36 kJ mol⁻¹

   Answer: C. 53.6 kJ mol⁻¹

   Explanation: The Arrhenius equation in two-temperature form gives ln(k₂/k₁) = (Ea/R)(T₂ − T₁)/(T₁T₂). Here ln 2 = 0.693 and (T₂ − T₁)/(T₁T₂) = 10/93000, so Ea = 0.693 × 8.314 × 9300 ≈ 53.6 × 10³ J mol⁻¹. Using log₁₀2 = 0.301 without inserting the 2.303 factor halves the answer to about 26.8 kJ mol⁻¹.

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