🔥🌟 MDCAT/ECAT/FSC Grand Chemistry MCQs Interactive Quiz 3 🧪💡| Learn & Practice!

1. If the Kelvin temperature of an ideal gas is increased to double and pressure is reduced to one-half, the volume of gas will:

(a) Remain same (b) Double (c) Half (d) Four times

✅ Correct Answer: (d) Four times
Doubling T → volume ×2; halving P → volume ×2; net effect = 4× increase.

2. Which of the following change is not an endothermic reaction?

(a) Cracking of alkanes (b) Decomposition of lime (c) Combustion of butane (d) Photosynthesis

✅ Correct Answer: (c) Combustion of butane
Combustion is exothermic (releases heat). The others absorb energy (endothermic).

3. For a hypothetical reaction x+y → z, if the conc. of x is double, the rate increases by square and if the conc. of y is doubled the rate increases by twice. The experimental rate law of this reaction is:

(a) R=K [x]¹ [y]¹ (b) R= K [x][y] (c) R=K [x][y] (d) R= K [x]² [y]¹

✅ Correct Answer: (d) R= K [x]² [y]¹

4. The unit of rate constant (K) for the zero order reaction is:

(a) s⁻¹ (b) conc. s⁻¹ (c) conc⁻¹. s (d) conc⁻¹. s⁻¹

✅ Correct Answer: (b) conc. s⁻¹

General rate law: R = k[A]ⁿ

where n = order of reaction.

Units of rate (R) = concentration/time (e.g., mol·L⁻¹·s⁻¹).

For a zero-order reaction (n = 0):

R = k[A]⁰ = k, So, K = R

Therefore, units of k = units of rate = concentration/time.

🔎 For zero-order reactions, the rate is independent of concentration, so the rate constant has the same units as the rate itself: concentration per unit time.

5. Which of the following is NOT a buffer solution?

(a) Na₂CO₃ /NaHCO₃ (b) NaOH/HCl (c) CH₃COOH/CH₃COONa (d) NH₄OH/NH₄Cl

✅ Correct Answer: (b) NaOH / HCl

Buffer solution: A mixture that resists changes in pH when small amounts of acid or base are added.

Typically made of a weak acid + its salt OR a weak base + its salt.

Check each option:

Na₂CO₃ / NaHCO₃→ Weak acid (HCO₃⁻) and its conjugate base (CO₃²⁻).✅ Buffer.

NaOH / HCl → Strong base + strong acid.❌ Not a buffer (they neutralize each other, no weak component remains).

CH₃COOH / CH₃COONa → Weak acid (acetic acid) + its salt (sodium acetate).✅ Buffer.

NH₄OH / NH₄Cl→Weak base (ammonium hydroxide) + its salt (ammonium chloride).✅ Buffer.

🔎 🔎 A buffer must involve a weak acid/base with its conjugate salt. NaOH and HCl are strong and neutralize completely, so they cannot form a buffer solution.

6. Conjugated acid of NH₃ is:

(a) NH₂⁻ (b) NH₄⁺ (c) NH₂/CH₃COONa (d) NH /NH₄Cl

✅ Correct Answer:(b) NH₄⁺

A conjugate acid is formed when a base accepts a proton (H⁺).

NH₃ (ammonia) acts as a base. If NH₃ accepts one proton:

NH₃ + H⁺ → NH₄⁺

So, the conjugate acid of NH₃ is NH₄⁺ (ammonium ion).

🔎 NH₃ is a weak base, and its conjugate acid is formed by adding one proton, giving NH₄⁺.

7. According to Graham’s Law of diffusion, the ratio of diffusion of H₂ and O₂ are respectively:

(a) 1:2 (b) 1:4 (c) 4:1 (d) 2:1

✅ Correct Answer:(c) 4:1

According to Graham's Law of Diffusion, the rate of diffusion of a gas is inversely proportional to the square root of its molar mass i.e. Rate of diffusion ∝ √1/M (M = molar mass of the gas).

For hydrogen (H₂) and oxygen (O₂):

Molar mass of H₂ = 2 g/mol

Molar mass of O₂ = 32 g/mol

Using Graham's Law:

Rate of diffusion of H₂/Rate of diffusion of O₂ = √32/2 = √16 = 4

Hence, the ratio of diffusion rates of H₂ to O₂ is 4:1.

🔎 According to Graham’s Law, lighter gases diffuse faster. Since H₂ is much lighter than O₂, it diffuses 4 times faster.

8. The molar volume of oxygen gas is maximum at:

(a) 0°C and 1 atm (b) 25°C and 1 atm (c) 0°C and 2 atm (d) 25°C and 2 atm

✅ Correct Answer: (b) 25°C and 1 atm

9. Under similar condition CH₄ gas diffuses........ times faster than SO₂ gas:

(a) 1.5 times (b) 4 times (c) 2 times (d) 16 times

✅ Correct Answer:(c) 2 times

10. A non-polar molecule with bigger size will experience:

(a) Hydrogen bonding (b) dipole-dipole interaction (c) London forces (d) All of these

✅ Correct Answer:(c) London forces

🔎 A non-polar molecule cannot form hydrogen bonds or dipole–dipole interactions. Its intermolecular attraction arises only from London dispersion forces, which become stronger as molecular size increases.

11. The temperature at which two allotropic forms co-exist in equilibrium is called as

(a) Melting temperature (b) fusion temperature (c) Transition temperature (d) critical temperature

✅ Correct Answer:(c) Hydrogen bond

🔎 When two allotropic forms of an element are in equilibrium, the specific temperature is known as the transition temperature. For example, rhombic sulfur converts to monoclinic sulfur at 95.6 °C (transition temperature).

12. Ice is a molecular solid, the intermolecular force of attraction among its molecules

(a) Ionic bond (b) Metallic bond (c) Hydrogen bond (d) Covalent bond

✅ Correct Answer:(c) Transition temperature

🔎 Ice is a molecular solid where water molecules are connected by hydrogen bonds, giving it its unique open lattice structure and lower density compared to liquid water. Covalent bonds exist within the H₂O molecule, but the intermolecular attraction that holds the lattice of ice together is hydrogen bonding.

13. If the bond angle is AB₂ type molecule is 104.5°, it geometry should be;

(a) Linear (b) Pyramidal (c) Bent (d) Planar Trigonal

✅ Correct Answer:(c) Bent

🔎 An AB₂ molecule with bond angle 104.5° (like H₂O) has 2 bond pairs and 2 lone pairs, giving it a bent geometry.

14. The shape and hybridization of BCl₃ molecule is:

(a) Tetrahedral and sp³ (b) Linear and sp (c) Angular & sp³ (d) Planar trigonal and sp²

✅ Correct Answer:(d) Planar trigonal and sp²

🔎 BCl₃ has 3 bond pairs, no lone pairs, giving a trigonal planar geometry with sp² hybridization.

15. Benzene has bond angles same as

(a) Tetrahedral geometry (b) Linear geometry (c) Triangular geometry (d) None of these

✅ Correct Answer:(c) Triangular geometry

🔎Benzene carbons are sp² hybridized, giving bond angles of 120°, the same as trigonal planar geometry.

16. Which one pair has same type of hybridization of central atom?

(a) BF₃, SO₃ (b) BF₃, NH₃ (c) CH₃⁺, BF₃ (d) Both a and c

✅ Correct Answers:(d) (a) BF₃, SO₃ → both sp² and (c) CH₃⁺, BF₃ → both sp²

(a) BF₃, SO₃: Both have sp² hybridization (trigonal planar geometry).

(b) BF₃, NH₃: BF₃ has sp² hybridization, while NH₃ has sp³ hybridization.

17. In O₂, each oxygen atom is …………. Hybridized.

(a) sp² (b) sp (c) sp³ (d) dsp³

✅ Correct Answers:(a) sp²

🔎In molecular oxygen (O₂), each oxygen atom forms one sigma bond and has two lone pairs.This corresponds to three electron domains, which means the hybridization is sp².

18. Bond length is largest in case of ………. overlaping

(a) sp²-s (b) sp²-sp² (c) sp³-sp³ (d) sp-sp

✅ Correct Answers:(c) sp³-sp³

🔎 Bond length is largest when orbitals have the least s-character. Since sp³ has only 25% s-character, sp³–sp³ overlap produces the longest bond.

19. In OF₂, number of bond pairs and lone pairs of electrons are respectively

(a) 2, 8 (b) 3, 1 (c) 2, 4 (d) 2, 2

✅ Correct Answers:(d) 2, 2

🔎 In OF₂, oxygen forms 2 bonds with fluorine and retains 2 lone pairs, giving a bent geometry.

20. Maximum how many numbers of hydrogen bond can be formed by H₂O molecule?

(a) 4 (b) 3 (c) 2 (d) 1

✅ Correct Answers:(a) 4

🔎 A water molecule can donate 2 hydrogen bonds (via its hydrogens) and accept 2 hydrogen bonds (via oxygen’s lone pairs). Hence, the maximum is 4.

21. The maximum number of electrons entering in a molecular orbital is:

(a) 2 (b) 3 (c) 4 (d) 1

✅ Correct Answers:(a) 1

🔎Each molecular orbital can accommodate a maximum of 2 electrons with opposite spins (Pauli Exclusion Principle), just like atomic orbitals.

22. Maximum number of molecules present in the following sample of gas:

(a) 100 g O₂ (b) 100 g Cl₂ (c) 100 g CO₂ (d) 100 g CH₄

✅ Correct Answer: (d) 100 g CH₄

(a) 100 g O₂ (Molar mass = 32 g/mol); n = 100/32 = 3.125 mol

(b) 100 g Cl₂ (Molar mass = 71 g/mol); n=100/71 = 1.41 mol

(d) 100 g CH₄ (Molar mass = 16 g/mol); n=100/16 = 6.25 mol

🔎 Since CH₄ has the lowest molar mass (16 g/mol), 100 g of it contains the largest number of moles and molecules.

23. This gas has the highest rate of diffusion at a given temperature and pressure

(a) Cl₂ (b) O₂ (c) F₂ (d) N₂

✅ Correct Answer:(d) N₂

🔎 Rate of diffusion ∝ 1 / √Molar mass → lighter gases diffuse faster.

Cl₂ → heaviest (Molar mass=71 g/mol) → slowest diffusion.

O₂ → 32 g/mol (Molar mass=32 g/mol) → moderate.

F₂ → 38 g/mol (Molar mass=38 g/mol) → slower than O₂.

N₂ → 28 g/mol (Molar mass=28 g/mol) → lightest → fastest diffusion. ✅

🔎 According to Graham’s Law, lighter gases diffuse faster. Since N₂ has the lowest molar mass (28 g/mol) among the options, it diffuses the fastest.

24. This principle involves in the liquefaction of gas

(a) Joule-Thomson effect (b) Henry’s Law (c) Le-Chatelier’s Principle (d) Raoult’s Law

✅ Correct Answers:(a) Joule-Thomson effect

🔎 Liquefaction of gases is based on the Joule–Thomson effect, where expansion without heat exchange causes cooling, enabling gases to condense into liquids.

25. One mole of H₂O contains this number of hydrogen atoms

(a) 3.01 × 10²³ (b) 6.02 × 10²³ (c) 1.204 × 10²⁴ (d) 1.204 × 10²³

✅ Correct Answer: (c) 1.204×10²⁴

1 mole of H₂O molecules = 6.022×10²³ molecules (Avogadro’s number).

Each H₂O molecule contains 2 hydrogen atoms.

Therefore, total hydrogen atoms in 1 mole of H₂O = mole × Nᴀ × atomicity =

1 × 2 × 6.022 × 10²³ = 1.204 × 10²⁴

🔎 Since each water molecule has 2 hydrogens, one mole of water contains twice Avogadro’s number of hydrogen atoms.

26. Amongst the following molecules which one has trigonal pyramidal shape?

(a) PCl₃ (b) SO₃²⁻ (c) NH₃ (d) All of them

✅ Correct Answers:(d) All of them

🔎 All three molecules (PCl₃, SO₃²⁻, NH₃) have 3 bond pairs and 1 lone pair on the central atom, giving a trigonal pyramidal geometry.

27. A simple covalent molecule possesses 3 bond pairs & 1 lone pair around the central atom, its shape should be:

(a) Pyramidal (b) Planar trigonal (c) Angular (d) Tetrahedral

✅ Correct Answers:(a) Pyramidal

🔎 With 3 bond pairs and 1 lone pair, the electron geometry is tetrahedral, but the molecular shape is pyramidal due to lone pair repulsion.

28. The correct relation between Debye and coulomb meter is:

(a) 1D = 1.88×10⁻¹²Cm (b) 1D = 1.6×10⁻¹⁹Cm (c) 1D = 3.33×10⁻³⁰Cm (d) 1D = 1.23×10⁻⁸Cm

✅ Correct Answers:(c) 1D = 3.33 × 10⁻³⁰Cm

🔎 The Debye is a small unit of dipole moment, and in SI units it equals 3.33 × 10⁻³⁰coulomb·meter.

29. The bond order of N₂ molecule is:

(a) 2 (b) 3 (c) 1 (d) 0

✅ Correct Answers:(b) 3

🔎 N₂ has a triple bond (one σ + two π bonds), giving a bond order of 3, which explains its high stability and inertness.

30. One mole of NaCl contains:

(a) 6.022 × 10²³ ions (b) 1.204 × 10²⁴ ions (c) 2 × 6.022 × 10²³ ions (d) Both (b) and (c)

✅ Correct Answers:(d) Both (b) and (c)

🔎 1 mole NaCl = 1 mole Na⁺ + 1 mole Cl⁻ = 2 × 6.022 × 10²³ = 1.204 × 10²⁴ ions.

31. The number of carbon atoms in 1 mole of glucose (C₆H₁₂O₆) are approximately:

(a) 6 × 10²³ (b) 36 × 10²³ (c) 60 × 10²³ (d) 72 × 10²³

✅ Correct Answers:(b) 36 × 10²³

🔎 Number of C atoms = n × Nᴀ × atomicity = 1 × 6.022 × 10²³ × 6 = 36.132 × 10²³ ≈ 36 × 10²³

32. Which contains the maximum number of atoms?

(a) 9 g H₂O (b) 6 g C (c) 7 g N₂ (d) 3 g He

✅ Correct Answers:(a) 9 g H₂O

🔎 No. of atoms in 9g H₂O (18 g)= n × Nᴀ × atomicity = 9/18 × Nᴀ × 3 = 1.5 Nᴀ atoms✅Maximum Number of atoms

🔎 No. of atoms in 6 g C (12 g) = n × Nᴀ × atomicity = 6/12 × Nᴀ × 1 = 0.5 Nᴀ atoms

🔎 No. of atoms in 7 g N₂ (14 g)= n × Nᴀ × atomicity = 7/14 × Nᴀ × 2 = 1 Nᴀ atoms

🔎 No. of atoms in 3 g He (4 g) = n × Nᴀ × atomicity = 3/4 × Nᴀ × 1 = 0.75 Nᴀ atoms

33. The prefix 10¹⁵

(a) Peta (b) Mega (c) Exa (d) Giga

✅ Correct Answers:(a) Peta

🔎 SI prefixes scale by powers of 10. Mega = 10⁶, Giga = 10⁹, Peta = 10¹⁵, and Exa = 10¹⁸. Since 10¹⁵matches peta, that’s the right prefix.

34. The mass of one mole of electrons is:

(a) 1.008 mg (b) 0.184 mg (c) 1.673 mg (d) 0.55 mg

✅ Correct Answers:(d) 0.55 mg

🔎 Mass of one electron is 9.11 × 10⁻³¹kg

Mass of one mole (6.02 × 10²³) electron is = 9.11 × 0⁻³¹× 6.02 × 10²³ = 54.84 × 10⁻⁸ kg or 5.5 × 10⁻⁷ kg

Mass of one mole of electron in mg = 5.5 × 10⁻⁷ kg × 1 × 106 = 5.5 × 10⁻¹ mg or 0.55 mg

35. The total number of ions in CaCl₂ is

(a) 6.02 × 10²³ (b) 12.04 × 10²³ (c) 18.06 × 10²³ (d) 24.08 × 10²³

✅ Correct Answers:(c) 18.06 × 10²³

🔎 Each mole of CaCl₂ produces 3 ions per formula unit, so the total number of ions = 3 × Avogadro’s number (3 × 6.02 × 10²³ = 18.06 × 10²³).

36. The value of azimuthal quantum number for the last electron of sodium atom is

(a) 3 (b) 2 (c) 1 (d) 0

✅ Correct Answers:(d) 0

🔎 Sodium (Na) has atomic number 11, so its electron configuration is 1s², 2s², 2p⁶, 3s¹

The last electron enters the 3s orbital.

For any s orbital, the azimuthal quantum number (ℓ) = 0

So the correct value of ℓ for the last electron of sodium is 0.

37. For which element would the valence shell electron configuration be ns²np⁵?

(a) Chlorine (Z=17) (b) Bromine (Z=35) (c) Iodine (Z=53) (d) All of them

✅ Correct Answers:(d) All of them

🔎 The valence configuration is ns²np⁵. This corresponds to halogens of group VIIA. All elements given are included in halogens. Fluorine is a first halogen having same valence shell electron configuration.

38. According to the Aufbau principle, the correct order for filling electrons is:

(a) 3d → 4s → 4p (b) 4s → 3d → 4p (c) 4p → 4s → 3d (d) 3d → 4p → 4s

✅ Correct Answers:(b) 4s → 3d → 4p

🔎 The Aufbau principle follows the order of increasing (n+l) value. If two subshells have the same (n+l) value, the one with the lower 'n' is filled first.

4s: n + ℓ = 4+0 = 4 (lowest n + ℓ) → filled first

3d: n + ℓ = 3+2 = 5

4p: n + ℓ = 4+1 = 5

So, 4s (lowest n + ℓ) is filled first. Then, between 3d and 4p (both n+ ℓ =5), 3d (lower n=3) is filled before 4p (n=4).

The order of filling electrons = 4s → 3d → 4p.

39. For n = 5, number of possible orbitals is:

(a) 50 (b) 10 (c) 25 (d) 50

✅ Correct Answers:(c) 25

🔎 The total number of orbitals for a given principal quantum number n is given by n².

Hence for n = 5, the number of possible orbitals is 5² i.e. 25

40. If an orbital has quantum numbers n = 4, l = 3, mₗ = –2, it belongs to:

(a) 4d (b) 4f (c) 5d (d) 5f

✅ Correct Answers:(b) 4f

🔎 n = 4 → Principal quantum number (energy level = 4)

ℓ = 3 → Azimuthal quantum number → f → So this is an f orbital

mₗ = –2 → Magnetic quantum number → Valid for ℓ = 3, since mₗ ranges from –3 to +3 (–3, –2, –1, 0, +1, +2, +3) → so –2 is valid

So the orbital is: 4f

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🔥🌟 MDCAT/ECAT/FSC Grand Chemistry MCQs Interactive Quiz 3 🧪💡| Learn & Practice!