JEE Chemistry Some Basic Concepts Of Chemistry MCQs Set 09

Answer: (a) 36 g of water

Explanation:
1. Moles of H₂O = 36/18 = 2
2. CO = 28/28 = 1
3. C₂H₅OH = 46/46 = 1
4. N₂O₅ = 54/108 = 0.5
5. The largest number of moles, 2, is in water, so it has the most molecules. Answer: (a)

Teacher's Note:
1. Number of molecules is proportional to the number of moles.
2. Convert each mass into moles using its molar mass.

Answer: (c) 7 : 32

Explanation:
1. Take 1 g of O₂ and 4 g of N₂
2. Moles of O₂ = 1/32 and moles of N₂ = 4/28 = 1/7
3. Ratio = (1/32) : (1/7) = 7 : 32. Answer: (c)

Teacher's Note:
1. Convert mass ratio into mole ratio by dividing by molar masses.
2. The ratio of molecules equals the ratio of moles.

Answer: (a) 24 g of C (atomic mass 12)

Explanation:
1. Moles of atoms: C = 24/12 = 2; Fe = 56/56 = 1; Al = 27/27 = 1; Ag = 108/108 = 1
2. Carbon has 2 mol, the largest, so it has the most atoms. Answer: (a)

Teacher's Note:
1. Number of atoms is proportional to moles of atoms.
2. Divide each mass by its atomic mass.

Answer: (b) 0.1 gram atom of Ca

Explanation:
1. Moles of CaCO₃ = 10/100 = 0.1
2. Each formula unit contains one Ca atom, so there is 0.1 gram atom of Ca
3. 0.1 mol is 6 × 10²² atoms, and 0.1 mol of Ca²⁺ is 0.2 equivalent. Answer: (b)

Teacher's Note:
1. Molar mass of CaCO3 is 100 g/mol.
2. A gram atom is simply one mole of atoms.

Answer: (a) 40, 30

Explanation:
1. Molar mass of XY₂ = 10/0.1 = 100, so \( X + 2Y = 100 \)
2. Molar mass of X₃Y₂ = 9/0.05 = 180, so \( 3X + 2Y = 180 \)
3. Subtracting, \( 2X = 80 \), so \( X = 40 \) and \( Y = 30 \). Answer: (a)

Teacher's Note:
1. Find each molar mass from mass divided by moles.
2. Subtract the two equations to eliminate Y.

Answer: (a) 0.060

Explanation:
1. Divide each amount by its coefficient: 0.10/4 = 0.025; 0.36/18 = 0.020; 0.90/30 = 0.030
2. SiO₂ gives the smallest value, so it is the limiting reagent
3. Moles of P₄ = 0.020 × 3 = 0.060. Answer: (a)

Teacher's Note:
1. The limiting reagent has the smallest moles-to-coefficient ratio.
2. Use that reagent to find the product amount.

Answer: (d) 106.4 kg

Explanation:
1. Pure CaCO₃ = 0.95 × 200 = 190 kg
2. \( \mathrm{CaCO_3 \to CaO + CO_2} \): 100 g gives 56 g of CaO
3. CaO = 190 × 56/100 = 106.4 kg. Answer: (d)

Teacher's Note:
1. Find the pure CaCO3 first.
2. Use the mass ratio 100 : 56 from the balanced equation.

Answer: (a) M/5

Explanation:
1. Equivalent weight = molar mass / number of electrons gained per formula unit
2. Mn changes from +7 to +2, so 5 electrons are gained
3. Equivalent weight = M/5. Answer: (a)

Teacher's Note:
1. In acidic medium permanganate takes up five electrons.
2. n-factor equals the change in oxidation number times number of atoms changing.

Answer: (a) 18.2, 32.69

Explanation:
1. Oxidant is V₂O₅: V goes from +5 to 0, so 10 electrons per formula unit
2. Molar mass of V₂O₅ = 2(50.94) + 80 = 181.88; equivalent weight = 181.88/10 = 18.2
3. Reductant is Zn: Zn goes from 0 to +2, so equivalent weight = 65.38/2 = 32.69. Answer: (a)

Teacher's Note:
1. The oxidant is reduced and the reductant is oxidised.
2. n-factor of V2O5 counts both vanadium atoms.

Answer: (d) Both (a) and (b)

Explanation:
1. H₃PO₃ is a dibasic acid, so on complete neutralisation to Na₂HPO₃ its equivalent weight is M/2
2. In partial neutralisation to NaH₂PO₃ only one H⁺ is replaced, so the equivalent weight is M
3. As a reducing agent, P goes from +3 to +5, giving M/2
4. So it may be M or M/2. Answer: (d)

Teacher's Note:
1. The equivalent weight of an acid depends on how many H+ ions take part in the reaction.
2. H3PO3 has two ionisable hydrogens, not three.

Answer: (b) 3

Explanation:
1. Per FeC₂O₄: Fe²⁺ → Fe³⁺ loses 1 electron and C₂O₄²⁻ → 2CO₂ loses 2 electrons, total 3 electrons
2. 5 mol lose 15 mol of electrons
3. Each KMnO₄ gains 5 electrons, so moles of KMnO₄ = 15/5 = 3. Answer: (b)

Teacher's Note:
1. Both the iron and the oxalate are oxidised.
2. Equate total electrons lost with total electrons gained.

Answer: (c) 6.32 g

Explanation:
1. Equivalents = 0.05 × 4 = 0.2
2. Equivalent weight of KMnO₄ in acidic medium = 158/5 = 31.6
3. Mass = 0.2 × 31.6 = 6.32 g. Answer: (c)

Teacher's Note:
1. In acidic medium the n-factor of KMnO4 is 5.
2. Mass equals equivalents times equivalent weight.

Answer: (b) 0.24 g

Explanation:
1. Equivalents of oxalic acid = 0.035 × 0.16 = 0.0056
2. Equivalents of MnO₂ reduced are the same, 0.0056
3. MnO₂ has n-factor 2 (Mn from +4 to +2), so equivalent weight = 87/2 = 43.5
4. Mass = 0.0056 × 43.5 = 0.2436 g, about 0.24 g. Answer: (b)

Teacher's Note:
1. At the end point, equivalents of oxidant equal equivalents of reductant.
2. Equivalent weight of MnO2 is half its molar mass.