Some Basic Concepts of Chemistry — JEE Main Questions

1. After balancing MnO4- + I- → MnO2 + I2 in basic medium, what is the coefficient of OH- on the product side?

1. 6
2. 4
3. 2
4. 8

Answer: 8

First balance in acidic medium: reduction: MnO4- + 4H+ + 3e- → MnO2 + 2H2O; oxidation: 2I- → I2 + 2e-. Multiply reduction by 2 and oxidation by 3: 2MnO4- + 8H+ + 6e- → 2MnO2 + 4H2O; 6I- → 3I2 + 6e-. Add: 2MnO4- + 6I- + 8H+ → 2MnO2 + 3I2 + 4H2O. Add 8OH- to both sides: 2MnO4- + 6I- + 8H2O → 2MnO2 + 3I2 + 4H2O + 8OH-. Cancel 4H2O: 2MnO4- + 6I- + 4H2O → 2MnO2 + 3I2 + 8OH-. So coefficient of OH- is 8.

2. 10 mL of a gaseous hydrocarbon is mixed with 100 mL of O2 (excess) and exploded. After cooling to room temperature (water condensed), the volume of the gaseous mixture is 80 mL. On passing this mixture through aqueous KOH, the volume reduces to 50 mL. What is the molecular formula of the hydrocarbon?

1. C2H6
2. C3H8
3. C3H6
4. C4H10

Answer: C3H8

Volume of CO2 produced = 80 − 50 = 30 mL (absorbed by KOH). Since 10 mL hydrocarbon gave 30 mL CO2, number of C atoms x = 30/10 = 3. Volume of O2 consumed = 100 − 50 = 50 mL (the 50 mL left after KOH is unreacted O2). For CxHy + (x + y/4) O2 → x CO2 + (y/2) H2O, O2 used per mL of hydrocarbon = 50/10 = 5, so x + y/4 = 5. With x = 3, y/4 = 2 ⇒ y = 8. Formula = C3H8.

3. In the balanced equation for MnO4- + I- → MnO2 + I2 in basic medium, what is the coefficient of OH-?

1. 8
2. 6
3. 4
4. 2

Answer: 8

Balance half-reactions: reduction: MnO4- + 4H+ + 3e- → MnO2 + 2H2O; oxidation: 2I- → I2 + 2e-. Multiply reduction by 2 and oxidation by 3, add: 2MnO4- + 6I- + 8H+ → 2MnO2 + 3I2 + 4H2O. Add 8OH- to both sides to neutralise H+: 2MnO4- + 6I- + 8H2O → 2MnO2 + 3I2 + 4H2O + 8OH-. Cancel 4H2O: 2MnO4- + 6I- + 4H2O → 2MnO2 + 3I2 + 8OH-. Coefficient of OH- is 8.

4. A 2.0 L flask contains 4.0 g of H2 and 32.0 g of O2 at 300 K. What is the partial pressure of H2 in atm? (R = 0.0821 L atm mol⁻¹ K⁻¹)

1. 12.3 atm
2. 4.1 atm
3. 24.6 atm
4. 8.2 atm

Answer: 24.6 atm

Using Dalton's law of partial pressures, partial pressure of H2 = mole fraction of H2 × total pressure. Moles of H2 = 4.0/2 = 2.0 mol; moles of O2 = 32.0/32 = 1.0 mol; total moles = 3.0 mol. Total pressure P_total = nRT/V = 3.0 × 0.0821 × 300 / 2.0 = 36.945 atm. Mole fraction of H2 = 2/3. Thus P_H2 = (2/3) × 36.945 = 24.63 atm ≈ 24.6 atm.

5. 0.50 g of impure CaCO3 is treated with 50.0 mL of 0.20 M HCl. The excess HCl requires 20.0 mL of 0.10 M NaOH for neutralisation. What is the percentage purity of CaCO3? (Molar mass CaCO3 = 100 g/mol)

1. 80%
2. 40%
3. 50%
4. 60%

Answer: 80%

Using neutralisation principle: initial moles HCl = 0.050 L × 0.20 M = 0.010 mol. Excess HCl = moles NaOH = 0.020 L × 0.10 M = 0.0020 mol. Moles HCl reacted = 0.010 - 0.0020 = 0.0080 mol. From CaCO3 + 2HCl → CaCl2 + CO2 + H2O, moles CaCO3 = 0.0080/2 = 0.0040 mol. Mass pure CaCO3 = 0.0040 × 100 = 0.40 g. Purity = (0.40/0.50) × 100 = 80%.

6. What is the coefficient of H+ in the balanced equation for MnO4- + Fe2+ + H+ → Mn2+ + Fe3+ + H2O?

1. 8
2. 5
3. 4
4. 10

Answer: 8

Using the oxidation-number method: Mn changes from +7 to +2 (gain of 5 e-), Fe changes from +2 to +3 (loss of 1 e-). Multiply Fe2+ by 5. Balance O: 4 O on left, so add 4 H2O on right. Balance H: 8 H on right, so add 8 H+ on left. The balanced equation is MnO4- + 5Fe2+ + 8H+ → Mn2+ + 5Fe3+ + 4H2O. Thus coefficient of H+ is 8.

7. Which of the following reactions is an example of comproportionation?

1. 2H2O2 → 2H2O + O2
2. 2NaHCO3 → Na2CO3 + CO2 + H2O
3. IO3- + 5I- + 6H+ → 3I2 + 3H2O
4. 2KClO3 → 2KCl + 3O2

Answer: IO3- + 5I- + 6H+ → 3I2 + 3H2O

In comproportionation, the same element in two different oxidation states reacts to form a single product in an intermediate oxidation state. In option b, iodine in IO3- (O.S. +5) and I- (O.S. -1) combine to give I2 (O.S. 0), which is an intermediate oxidation state. This matches the definition.

8. In an iodometric estimation, 25 mL of a CuSO4 solution liberated I2 which required 20 mL of 0.1 N hypo. What is the mass of copper in the sample? (Atomic mass of Cu = 63.5 g/mol)

1. 0.0635 g
2. 0.254 g
3. 0.127 g
4. 0.3175 g

Answer: 0.127 g

In the reaction 2Cu2+ + 4I- → Cu2I2 + I2, each Cu2+ gains one electron, so n-factor of Cu2+ is 1. The liberated I2 is titrated with hypo (n=1). By law of equivalence, meq of Cu = meq of hypo = N×V = 0.1×20 = 2 meq. Mass of Cu = (meq × equivalent mass)/1000 = (2 × 63.5)/1000 = 0.127 g.

9. When balancing Cr2O7^2- + C2O4^2- + H+ → Cr3+ + CO2 + H2O in acidic medium, what is the coefficient of H+ in the balanced equation?

1. 6
2. 8
3. 14
4. 7

Answer: 14

The reduction half-reaction is Cr2O7^2- + 14H+ + 6e- → 2Cr3+ + 7H2O. The oxidation half-reaction is C2O4^2- → 2CO2 + 2e-. Multiply oxidation by 3 to equalize electrons: 3C2O4^2- → 6CO2 + 6e-. Adding gives Cr2O7^2- + 3C2O4^2- + 14H+ → 2Cr3+ + 6CO2 + 7H2O. Thus coefficient of H+ is 14.

10. 5.4 g of Al and 10.65 g of Cl₂ are reacted according to 2 Al + 3 Cl₂ → 2 AlCl₃. Which is the limiting reagent?

1. Cl₂
2. Al
3. Both are limiting
4. Neither is limiting

Answer: Cl₂

Limiting reagent is determined by comparing the mole ratio of reactants to the stoichiometric coefficients. Moles of Al = 5.4/27 = 0.2; moles of Cl₂ = 10.65/71 = 0.15. Divide by coefficients: Al: 0.2/2 = 0.1; Cl₂: 0.15/3 = 0.05. The smaller quotient (0.05) indicates Cl₂ is limiting.

11. In a titration, 0.5 g of K2Cr2O7 (M = 294 g/mol) is dissolved in dilute H2SO4 and used to oxidise Fe2+ to Fe3+. How many milliequivalents of Fe2+ are oxidised?

Answer: 10.2

The n-factor of K2Cr2O7 in acidic medium is 6 (Cr2O7^2- + 14H+ + 6e- → 2Cr3+ + 7H2O). Number of equivalents of K2Cr2O7 = mass / (M/n) = 0.5 / (294/6) = 0.5 / 49 = 0.0102 eq = 10.2 meq. By law of equivalence, equivalents of Fe2+ = equivalents of K2Cr2O7 = 10.2 meq.

12. A mixture of Na2CO3 and NaHCO3 requires 20 mL of 0.1 M HCl with phenolphthalein and 50 mL of same HCl with methyl orange. What are the millimoles of Na2CO3 and NaHCO3?

1. Na2CO3 = 1 mmol, NaHCO3 = 4 mmol
2. Na2CO3 = 2 mmol, NaHCO3 = 3 mmol
3. Na2CO3 = 3 mmol, NaHCO3 = 2 mmol
4. Na2CO3 = 2.5 mmol, NaHCO3 = 2.5 mmol

Answer: Na2CO3 = 2 mmol, NaHCO3 = 3 mmol

Phenolphthalein titre (V1=20 mL) corresponds to half of Na2CO3: meq HCl = 0.1×20 = 2, so meq Na2CO3 = 2×2 = 4, mmol = 4/2 = 2. Methyl orange titre (V2=50 mL) gives total meq = 0.1×50 = 5. Bicarbonate from Na2CO3 = 2 meq, so meq NaHCO3 = 5-2 = 3, mmol = 3/1 = 3.