MDCAT/ECAT/FSC Grand Chemistry MCQs Quiz # 2| Interactive Quiz
1. Salt which is formed by the neutralization of weak acid and strong base is:
a
NaNO₃
Wrong
b
NH₄Cl
Wrong
c
Na₂CO₃
Right
d
NH₄CN
Wrong
Answer: (c)
Na₂CO₃ is formed from weak acid (H₂CO₃) and strong base (NaOH).
2. Which one of the following cannot have value equal to zero?
a
Activation energy
Wrong
b
Reduction potential
Wrong
c
Principal quantum number
Wrong
d
Both a and c
Right
Answer: (d)
Activation energy cannot be zero (needs minimum energy to start reaction).
Principal quantum number (n) starts from 1, cannot be zero.
3. Which of the following enthalpy change is always negative?
a
∆Hᵣ°
Wrong
b
∆Hf°
Wrong
c
∆Hc°
Right
d
∆Hd°
Wrong
Answer: (c)
∆Hc° (Enthalpy of Combustion) is always negative because combustion reactions are exothermic, meaning they release energy (heat) to the surroundings.
In contrast, other enthalpy changes such as ∆Hf° (standard enthalpy of formation) and ∆Hᵣ° (reaction enthalpy) can be either positive or negative depending on whether the process is exothermic or endothermic. ∆Hd° refers to enthalpy of decomposition, which can also be either exothermic or endothermic.
4. The energy corresponds to the given thermochemical process is labeled as:
Li⁺ (g) + Cl⁻ (g) → LiCl(s)
a
Ionization energy
Wrong
b
Enthalpy of atomization
Wrong
c
Enthalpy of combustion
Wrong
d
Lattice energy
Right
Answer: (d)
Lattice energy is the energy released when gaseous ions (Li⁺ and Cl⁻) combine to form an ionic solid (LiCl). It reflects the strength of the ionic bond formed in the solid lattice.
5. In the equation of First law of thermodynamics (∆E=q + w), the property(s) which does not depend upon initial and final state is (are):
a
Ionization energy
Wrong
b
W
Wrong
c
q
Wrong
d
Both q and W
Right
Answer: (d)
∆E (change in internal energy) is a state function, meaning it depends only on the initial and final states of the system, not on the path taken.
q (heat) and W (work) are path functions, meaning they depend on the process or path taken to go from the initial to the final state. Therefore, q and W do not only depend on the initial and final states, but on how the process occurs.
Thus, q and W depend on the path, not just the initial and final states, making them the correct answer.
6. The strongest oxidizing agent and strongest reducing agent in the electro chemical series respectively is:
a
Li and F
Wrong
b
Li and H₂
Wrong
c
F and Li
Right
d
H₂ and F
Wrong
Answer: (c)
F₂ (Fluorine) is the strongest oxidizing agent because it has the highest tendency to gain electrons and get reduced (as it has a very high reduction potential).
Li (Lithium) is the strongest reducing agent because it has the highest tendency to lose electrons and get oxidized (it has the lowest reduction potential).
Thus, F (as an oxidizing agent) and Li (as a reducing agent) are the strongest in their respective categories in the electrochemical series.
7. Least entropy found in which of the following state of water:
a
Steam at 100°C
Wrong
b
Liquid water at 25°C
Wrong
c
Liquid water at 4°C
Wrong
d
Ice at 0°C
Right
Answer: (d)
Entropy is a measure of disorder or randomness. The least entropy is found in the solid state of water (ice) because the molecules are arranged in an orderly structure, leading to minimal disorder.
As water moves from solid (ice) to liquid and then to gas (steam), the entropy increases due to more freedom of movement and disorder in the molecular arrangement.
Therefore, ice at 0°C has the least entropy because the molecules are in a structured, fixed lattice, resulting in the lowest degree of disorder.
8. Which of the following possesses weakest London dispersion forces:
a
Cl₂
Wrong
b
F₂
Right
c
Br₂
Wrong
d
I₂
Wrong
Answer: (b)
London dispersion forces (LDF) increase with the size and polarizability of the molecule.
F₂ is the smallest molecule with the least number of electrons among the options, resulting in the weakest dispersion forces.
As the size of the halogen molecule increases (Cl₂, Br₂, I₂), the number of electrons increases, which leads to stronger London dispersion forces.
9. In NaCl, each Na ion is surrounded by CI ions in the numbers:
a
Four
Wrong
b
Three
Wrong
c
Six
Right
d
Seven
Wrong
Answer: (c)
In NaCl (sodium chloride), the structure is a face-centered cubic (FCC) lattice. Each Na⁺ ion is surrounded by 6 Cl⁻ ions in an octahedral arrangement, and vice versa. This means each ion in NaCl is coordinated to 6 oppositely charged ions.
10. Oxidation number of Cr in Na₂Cr₂O₇ is:
a
+6
Right
b
+12
Wrong
c
+7
Wrong
d
+5
Wrong
Answer: (a)
In Na₂Cr₂O₇, the total oxidation number of oxygen is -14 and sodium is +2. To balance this, each chromium (Cr) has an oxidation number of +6.
11. The number of bonds in ethyne (C₂H₂) is:
a
2 sigma, 2 pi
Wrong
b
3 sigma, two pi
Right
c
3 sigma, 3 pi
Wrong
d
2 sigma, 3 pi
Wrong
Answer: (b)
Ethyne has 1 C≡C triple bond (1 σ + 2 π) and 2 C–H single bonds (2 σ), giving a total of 3 σ and 2 π bonds.
12. A colloidal solution of liquid into liquid is known as:
a
Gel
Wrong
b
Foam
Wrong
c
Sol
Wrong
d
Emulsion
Right
Answer: (d)
A colloidal solution is a mixture where one substance is dispersed in another.
When liquid is dispersed in another liquid, it forms an emulsion.
Examples: milk (fat in water), mayonnaise (oil in water).
Other options:
Gel: solid in liquid
Foam: gas in liquid
Sol: solid in liquid
So, the correct term for liquid in liquid colloid is Emulsion.
13. VBT tells us about all of the following facts except:
a
Bond strength
Wrong
b
Bond length
Wrong
c
Bond order
Wrong
d
Bond energy
Right
Answer: (d)
Valence Bond Theory (VBT) explains:
Bond formation through overlapping atomic orbitals
Bond strength (qualitative)
Bond length (based on orbital overlap)
Bond order (number of shared electron pairs)
However, bond energy (quantitative energy required to break a bond) is not predicted by VBT; it is explained by Molecular Orbital Theory or experimental data.
14. The unit of rate constant (K) for the first order reaction is:
a
s⁻¹
Right
b
conc. s⁻¹
Wrong
c
conc⁻¹. s
Wrong
d
conc⁻¹. s⁻¹
Wrong
Answer: (a)
For a first-order reaction, the rate law is: Rate=k[A]
Rate has units of concentration × time⁻¹ (e.g., mol·L⁻¹·s⁻¹)
[A] has units of concentration (mol·L⁻¹)
To find k, you divide the units of rate by the units of [A]:
k units=(concentration × time⁻¹) per concentration=time⁻¹
✅ Conclusion: The rate constant k for a first-order reaction has units of s⁻¹.
15. The decomposition of H₂O₂ is inhibited by:
a
2% ethanol
Wrong
b
Glycerin
Right
c
MnO₂
Wrong
d
V₂O₅
Wrong
Answer: (b)
H₂O₂ naturally decomposes into H₂O and O₂.
Its decomposition is catalyzed by MnO₂ or V₂O₅.
Inhibitors like glycerin/glycerol slow down decomposition by stabilizing H₂O₂.
✅ Conclusion: Glycerin inhibits the decomposition of H₂O₂.
16. The sum of mole fractions of components of a solution is equal to:
a
0.0
Wrong
b
1.0
Right
c
10
Wrong
d
100
Wrong
Answer: (b)
Mole fraction (Xᵢ) of a component = moles of that component ÷ total moles of all components.
For a solution with multiple components:
X₁ + X₂ + X₃ +… = 1
✅ Conclusion: The sum of mole fractions of all components in a solution is always 1.0.
17. Conjugate base and conjugate acid of HCO₃⁻ are respectively
a
CO₃²⁻ and H₂CO₃
Right
b
H₂CO₃ and CO₃²⁻
Wrong
c
H⁺ and H₂O
Wrong
d
CO₃²⁻ and H₂O
Wrong
Answer: (a)
Amphiprotic: Can act as acid or base e.g. HCO₃⁻
As acid: HCO₃⁻ → loses H⁺ → CO₃²⁻ (conjugate base)
As base: HCO₃⁻ → gains H⁺ → H₂CO₃ (conjugate acid)
✅ Quick Tip: “HCO₃⁻ sits in the middle: one step up → CO₃²⁻, one step down → H₂CO₃.”
18. This liquid has the highest surface tension :
a
Mercury
Right
b
Water
Wrong
c
Ethanol
Wrong
d
Gasoline
Wrong
Answer: (a)
Mercury has the highest surface tension because of its strong cohesive metallic bonding between atoms. These forces are much stronger than hydrogen bonding in water or weak van der Waals forces in ethanol and gasoline, giving mercury the maximum surface tension among common liquids.
Typical values at room temperature (approximate):
Mercury: ~485 mN/m (very high, due to strong metallic bonding and cohesion)
Water: ~72 mN/m
Ethanol: ~22 mN/m
Gasoline: ~20–25 mN/m
👉 Clearly, Mercury has the highest surface tension among these liquids.
19. In equilibrium system; PCl₅ ⇌ PCl₃ + Cl₂; The relationship between Kₚ and Kc is :
a
Kₚ < Kc
Wrong
b
Kₚ › Kc
Right
c
Kₚ = Kc
Wrong
d
Kₚ≠ Kc
Wrong
Answer: (b)
The general relationship between Kₚ and Kc:
Kₚ =Kc(RT)▵ⁿ
Here, Δn = moles of gaseous products − moles of gaseous reactants.
For the reaction: PCl₅ (g)⇌PCl₃ (g)+Cl₂ (g); Δn = (1 + 1) − 1 = 1
Since Δn > 0 → Kₚ = Kc × (RT)¹ → Kₚ > Kc
✅ Conclusion: For reactions where the number of gaseous molecules increases, Kₚ is greater than Kc.
20. The extent of reaction will be maximum for the Kc value
a
0.1
Wrong
b
10
Wrong
c
10³
Right
d
10⁻³
Wrong
Answer: (c)
💡 Extent of reaction increases with larger equilibrium constant (Kc).
A large Kc (≫1) → almost complete reaction (products favored)
A small Kc ≪ 1 → reaction barely proceeds (reactants favored)
Here, Kc = 10³ is the largest, so maximum reaction extent occurs.
21. If the radius of Zn²⁺ ion is 0.74Å and that of S²⁻ ion is 1.84 Å, the radius ratio of ZnS should be:
a
2.48
Wrong
b
1.84
Wrong
c
0.74
Wrong
d
0.40
Right
Answer: (d)
Radius ratio (r⁺/r⁻) = radius of cation ÷ radius of anion rZn²⁺/rS²⁻= 0.74 Å /1.84 Å ≈ 0.40 ✅ Conclusion: The radius ratio of ZnS is 0.40. 💡 Tip: Radius ratio helps predict ionic crystal structures. 0.225–0.414 → tetrahedral 0.414–0.732 → octahedral
Radius ratio (r⁺/r⁻) = radius of cation ÷ radius of anion rZn²⁺/rS²⁻= 0.74 Å /1.84 Å ≈ 0.40 ✅ Conclusion: The radius ratio of ZnS is 0.40. 💡 Tip: Radius ratio helps predict ionic crystal structures. 0.225–0.414 → tetrahedral 0.414–0.732 → octahedral
22. Milk is an example of this type of colloid
a
Gel
Wrong
b
Aerosol
Wrong
c
Foam
Wrong
d
Emulsion
Right
Answer: (d)
Milk is a liquid-in-liquid colloid, where fat droplets are dispersed in water. Colloids with one liquid dispersed in another are called emulsions. Other types for reference: Gel: solid in liquid Aerosol: liquid/gas dispersed in gas Foam: gas in liquid ✅ Conclusion: Milk is an emulsion.
Milk is a liquid-in-liquid colloid, where fat droplets are dispersed in water. Colloids with one liquid dispersed in another are called emulsions. Other types for reference: Gel: solid in liquid Aerosol: liquid/gas dispersed in gas Foam: gas in liquid ✅ Conclusion: Milk is an emulsion.
23. Significant figures in 0.0880 are
a
2
Wrong
b
5
Wrong
c
4
Wrong
d
3
Right
Answer: (d)
0.0880 → Leading zeros do not count All non-zero digits count Trailing zero after decimal counts Digits counted: 8, 8, 0 → 3 significant figures
0.0880 → Leading zeros do not count All non-zero digits count Trailing zero after decimal counts Digits counted: 8, 8, 0 → 3 significant figures
24. The number of orbitals in fourth energy level is
a
4
Wrong
b
9
Wrong
c
32
Wrong
d
16
Right
Answer: (d)
Number of orbitals in a shell = n², where n = principal quantum number. For n = 4 → 4² = 16 orbitals 💡 Tip: Each orbital can hold 2 electrons, so maximum electrons in 4th shell = 16 × 2 = 32.
Number of orbitals in a shell = n², where n = principal quantum number. For n = 4 → 4² = 16 orbitals 💡 Tip: Each orbital can hold 2 electrons, so maximum electrons in 4th shell = 16 × 2 = 32.
25. The number of sigma and pi bonds in C₂H₆ (ethane) are
a
7 and none
Right
b
7 and 1
Wrong
c
3 and 3
Wrong
d
3 and 2
Wrong
Answer: (a)
C₂H₆ (ethane) is a single-bonded molecule: H₃C–CH₃ Sigma (σ) bonds: C–C single bond → 1 σ C–H bonds → 6 σ Total σ bonds = 7 Pi (π) bonds: No double or triple bonds → 0 π bonds ✅ Conclusion: 7 sigma bonds and no pi bonds in C₂H₆.
C₂H₆ (ethane) is a single-bonded molecule: H₃C–CH₃ Sigma (σ) bonds: C–C single bond → 1 σ C–H bonds → 6 σ Total σ bonds = 7 Pi (π) bonds: No double or triple bonds → 0 π bonds ✅ Conclusion: 7 sigma bonds and no pi bonds in C₂H₆.
26. This molecule has minimum bond angle
a
CS₂
Wrong
b
BF₃
Wrong
c
H₂O
Right
d
NH₃
Wrong
Answer: (c)
💡 Bond angle decreases with increasing lone pair repulsion. Approximate bond angles: CS₂: 180° (linear) BF₃: 120° (trigonal planar) NH₃: 107° (trigonal pyramidal) H₂O: 104.5° (bent) → minimum bond angle ✅ Conclusion: H₂O has the smallest bond angle due to two lone pairs on oxygen.
💡 Bond angle decreases with increasing lone pair repulsion. Approximate bond angles: CS₂: 180° (linear) BF₃: 120° (trigonal planar) NH₃: 107° (trigonal pyramidal) H₂O: 104.5° (bent) → minimum bond angle ✅ Conclusion: H₂O has the smallest bond angle due to two lone pairs on oxygen.
27. The geometry NH₄⁺ and SO₄²⁻ of ions is
a
Trigonal
Wrong
b
Pyramidal
Wrong
c
Tetrahedral
Right
d
Square planar
Wrong
Answer: (c)
💡 NH₄⁺ (ammonium ion): central N atom bonded to 4 H atoms → tetrahedral geometry SO₄²⁻ (sulfate ion): central S atom bonded to 4 O atoms → tetrahedral geometry ✅ Conclusion: Both NH₄⁺ and SO₄²⁻ ions have tetrahedral geometry.
💡 NH₄⁺ (ammonium ion): central N atom bonded to 4 H atoms → tetrahedral geometry SO₄²⁻ (sulfate ion): central S atom bonded to 4 O atoms → tetrahedral geometry ✅ Conclusion: Both NH₄⁺ and SO₄²⁻ ions have tetrahedral geometry.
28. This pair do not obey Dalton’s law:
a
He and H₂
Wrong
b
H₂ and Ar
Wrong
c
HCl and NH₃
Right
d
He and NH₃
Wrong
Answer: (c)
💡 Dalton’s Law of Partial Pressures applies to gases that do not react. HCl and NH₃ react to form NH₄Cl (solid) → gas mixture is not ideal Other pairs (He & H₂, H₂ & Ar, He & NH₃) do not react → obey Dalton’s Law ✅ Conclusion: HCl and NH₃ do not obey Dalton’s Law.
💡 Dalton’s Law of Partial Pressures applies to gases that do not react. HCl and NH₃ react to form NH₄Cl (solid) → gas mixture is not ideal Other pairs (He & H₂, H₂ & Ar, He & NH₃) do not react → obey Dalton’s Law ✅ Conclusion: HCl and NH₃ do not obey Dalton’s Law.
29. The vapour pressure of H₂O at 100°C is
a
76 torr
Wrong
b
5 atm
Wrong
c
101325 pascal
Right
d
12.5 psi
Wrong
Answer: (c)
💡 At 100°C, water boils at 1 atm pressure. 1 atm = 101325 Pa = 760 mmHg (torr) So, the vapor pressure of H₂O at 100°C is 101325 pascal ✅ Conclusion: Boiling occurs when vapor pressure = atmospheric pressure.
💡 At 100°C, water boils at 1 atm pressure. 1 atm = 101325 Pa = 760 mmHg (torr) So, the vapor pressure of H₂O at 100°C is 101325 pascal ✅ Conclusion: Boiling occurs when vapor pressure = atmospheric pressure.
30. Diamond is an example of
a
Ionic solid
Wrong
b
Molecular solid
Wrong
c
Covalent solid
Right
d
Metallic solid
Wrong
Answer: (c)
💡 Diamond is made of carbon atoms, each covalently bonded to four other carbon atoms in a 3D network. It has strong covalent bonds throughout, forming a giant covalent structure. ✅ Conclusion: Diamond is a covalent (network) solid.
💡 Diamond is made of carbon atoms, each covalently bonded to four other carbon atoms in a 3D network. It has strong covalent bonds throughout, forming a giant covalent structure. ✅ Conclusion: Diamond is a covalent (network) solid.
31. If a≠ b ≠c and α = γ = 90° and β ≠ 90°, then the crystal structure is
a
Monoclinic
Right
b
Orthorhombic
Wrong
c
Triclinic
Wrong
d
Hexagonal
Wrong
Answer: (a)
In a monoclinic crystal system: Edge lengths: a ≠ b ≠ c Angles: α = γ = 90°, β ≠ 90° Other systems for reference: Orthorhombic: a ≠ b ≠ c, α = β = γ = 90° Triclinic: a ≠ b ≠ c, α ≠ β ≠ γ ≠ 90° Hexagonal: a = b ≠ c, α = β = 90°, γ = 120° ✅ Conclusion: The given parameters describe a monoclinic crystal.
In a monoclinic crystal system: Edge lengths: a ≠ b ≠ c Angles: α = γ = 90°, β ≠ 90° Other systems for reference: Orthorhombic: a ≠ b ≠ c, α = β = γ = 90° Triclinic: a ≠ b ≠ c, α ≠ β ≠ γ ≠ 90° Hexagonal: a = b ≠ c, α = β = 90°, γ = 120° ✅ Conclusion: The given parameters describe a monoclinic crystal.
32. This oxide is amphoteric:
a
K₂O
Wrong
b
Al₂O₃
Right
c
CO₂
Wrong
d
CaO
Wrong
Answer: (b)
Amphoteric oxides react with both acids and bases. Al₂O₃ is amphoteric as Al₂O₃ reacts with both acid and base: Acid: Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O Base: Al₂O₃ + 2NaOH + 3H₂O → 2Na[Al(OH)₄] Other oxides: K₂O, CaO → basic oxides CO₂ → acidic oxide
Amphoteric oxides react with both acids and bases. Al₂O₃ is amphoteric as Al₂O₃ reacts with both acid and base: Acid: Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O Base: Al₂O₃ + 2NaOH + 3H₂O → 2Na[Al(OH)₄] Other oxides: K₂O, CaO → basic oxides CO₂ → acidic oxide
33. If the rate law expression is R = K, its order of reaction is
a
First
Wrong
b
Second
Wrong
c
Third
Wrong
d
Zero
Right
Answer: (d)
Rate law: R=K Rate does not depend on concentration of reactants → zero order reaction General form for zero order: R=k[A]⁰ =k ✅ Conclusion: Order of reaction is zero.
Rate law: R=K Rate does not depend on concentration of reactants → zero order reaction General form for zero order: R=k[A]⁰ =k ✅ Conclusion: Order of reaction is zero.
34. No. of atoms in 60g carbon will be:
a
3.01×10²⁴
Right
b
3.0 ×10²³
Wrong
c
6.02×10²³
Wrong
d
6.02 ×10²⁴
Wrong
Answer: (a)
Molar mass of carbon = 12 g/mol Moles of C in 60 g = 60 ÷ 12 = 5 mol Number of atoms = moles × Avogadro’s number 5 × 6.02 × 10²³ = 3.01×10²⁴ atoms
Molar mass of carbon = 12 g/mol Moles of C in 60 g = 60 ÷ 12 = 5 mol Number of atoms = moles × Avogadro’s number 5 × 6.02 × 10²³ = 3.01×10²⁴ atoms
35. Which of the following does NOT alter the pH of a solution?
a
NH₄Cl
Wrong
b
Na₂CO₃
Wrong
c
NaCl
Right
d
Mg(OH)Cl
Wrong
Answer: (c)
💡 NaCl is a neutral salt (strong acid HCl + strong base NaOH) → does not hydrolyze → pH remains ~7 Other salts: NH₄Cl → acidic solution (NH₄⁺ hydrolyzes) Na₂CO₃ → basic solution (CO₃²⁻ hydrolyzes) Mg(OH)Cl → slightly basic/acidic depending on hydrolysis
💡 NaCl is a neutral salt (strong acid HCl + strong base NaOH) → does not hydrolyze → pH remains ~7 Other salts: NH₄Cl → acidic solution (NH₄⁺ hydrolyzes) Na₂CO₃ → basic solution (CO₃²⁻ hydrolyzes) Mg(OH)Cl → slightly basic/acidic depending on hydrolysis
36. The unit of rate constant (K) for the first order reaction is:
a
conc. s⁻¹
Wrong
b
conc⁻¹. s
Wrong
c
s⁻¹
Right
d
conc⁻¹. s⁻¹
Wrong
Answer: (c)
💡 First-order reaction: Rate = k × [A] Rate has units of concentration × time⁻¹ [A] has units of concentration k units = time⁻¹ = s⁻¹ ✅ Conclusion: The rate constant k for a first-order reaction has units s⁻¹.
💡 First-order reaction: Rate = k × [A] Rate has units of concentration × time⁻¹ [A] has units of concentration k units = time⁻¹ = s⁻¹ ✅ Conclusion: The rate constant k for a first-order reaction has units s⁻¹.
37. The mole fractions of component A in a solution is 0.3. The mole fraction of component B is
a
1.0
Wrong
b
0.3
Wrong
c
0.5
Wrong
d
0.7
Right
Answer: (d)
💡 Sum of mole fractions in a solution = 1 Mole fraction of B = 1 − mole fraction of A = 1 − 0.3 = 0.7
💡 Sum of mole fractions in a solution = 1 Mole fraction of B = 1 − mole fraction of A = 1 − 0.3 = 0.7
38. The conduction of electricity through an electrolytic solution is due to the flow of
a
Electrons
Wrong
b
Atoms
Wrong
c
Molecules
Wrong
d
ions
Right
Answer: (d)
💡 In electrolytic solutions, ions are present and carry electric charge. Cations move toward the cathode and anions move toward the anode, producing current. ✅ Conclusion: Conduction of electricity in an electrolytic solution is due to ion movement.
💡 In electrolytic solutions, ions are present and carry electric charge. Cations move toward the cathode and anions move toward the anode, producing current. ✅ Conclusion: Conduction of electricity in an electrolytic solution is due to ion movement.
39. At 4°C the density of water is 1 g/cm³ then the mass of 2dm³ of water is
a
18 g
Wrong
b
100 g
Wrong
c
1000 g
Wrong
d
2000 g
Right
Answer: (d)
💡 Density of water: 1 g/cm³ = 1 g/mL Volume given: 2 dm³ = 2000 mL = 2000 cm³ Mass = Density × Volume = 1 × 2000 = 2000 g ✅ Conclusion: Mass of 2 dm³ of water = 2000 g.
💡 Density of water: 1 g/cm³ = 1 g/mL Volume given: 2 dm³ = 2000 mL = 2000 cm³ Mass = Density × Volume = 1 × 2000 = 2000 g ✅ Conclusion: Mass of 2 dm³ of water = 2000 g.
40. Total number of atoms in one mole of methane (CH₄) is:
a
6.02 × 10²³
Wrong
b
3.01 × 10²⁴
Right
c
1.81 × 10²³
Wrong
d
30.1 × 10²⁴
Wrong
Answer: (b)
💡 CH₄ has 1 C + 4 H = 5 atoms per molecule 1 mole of CH₄ = 6.02 × 10²³ molecules Total atoms = 6.02 × 10²³ × 5 ≈ 3.01 × 10²⁴ atoms ✅ Conclusion: One mole of methane contains 3.01 × 10²⁴ atoms.
💡 CH₄ has 1 C + 4 H = 5 atoms per molecule 1 mole of CH₄ = 6.02 × 10²³ molecules Total atoms = 6.02 × 10²³ × 5 ≈ 3.01 × 10²⁴ atoms ✅ Conclusion: One mole of methane contains 3.01 × 10²⁴ atoms.
Final Score
🔥 جونؔ ایلیا 🔥
تاریخ کی قبروں میں اُتر کیوں نہیں جاتے ⚰️
یہ گزرے ہوئے لوگ، گزر کیوں نہیں جاتے 🕰️
بازار میں بیزار ہی جب پھرتے ہو خود سے 🛍️
پھر گھر میں تو وحشت بھی ہے گھر کیوں نہیں جاتے 🏠
ہے شہر میں کیا بات بیاباں سے زیادہ 🌆
دیوانے بہت دن سے اُدھر کیوں نہیں جاتے 🏜️
کرنی ہے عطا تم کو اگر خواہشِ پرواز 🕊️
تم آکے پر و بال کتر کیوں نہیں جاتے ✈️
ہے دین کے اس دور میں دانش جنھیں پیاری 📖
وہ بگڑے ہوئے لوگ سُدھر کیوں نہیں جاتے 🔄
جو لوگ کہ اس دور کا آزار ہیں وہ لوگ ⚡
مارے نہیں جاتے ہیں تو مر کیوں نہیں جاتے 💀
مومن ہو تو بے تیغ بھی لڑتا ہے سپاہی ⚔️
میدان میں بے تیغ و سپر کیوں نہیں جاتے 🛡️
Tags
Aptitude test (MDCAT/ECAT)
chemistry learning portal
chemistry MCQs 2026
First year
FSC chemistry practice
Interactive Chemistry Questions
MDCAT ECAT Chemistry MCQs
online chemistry quiz
physics chemistry