Some Basic Concepts of Chemistry Practice

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NEET Chemistry - Chapter 1

Some Basic Concepts of Chemistry Practice Tests

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Mole idea
1. One mole of any substance contains:
- A. $6.022\times10^{23}$ entities (Correct)
- B. $3.011\times10^{23}$ entities
- C. 1 gram of particles
- D. 22.4 particles
Explanation: One mole is defined by Avogadro constant.
Mole idea
2. The mass of one mole of oxygen atoms is:
- A. 16 g (Correct)
- B. 8 g
- C. 32 g
- D. 1 g
Explanation: Atomic mass of oxygen is 16 u, so molar mass of O atoms is 16 g mol$^{-1}$.
Mole idea
3. The number of moles in 11 g of $CO_2$ is:
- A. 0.25 (Correct)
- B. 0.5
- C. 1
- D. 2
Explanation: Molar mass of $CO_2$ is 44 g mol$^{-1}$, so moles = 11/44 = 0.25.
Molar mass
4. Molar mass of $CaCO_3$ is:
- A. 100 g mol$^{-1}$ (Correct)
- B. 40 g mol$^{-1}$
- C. 56 g mol$^{-1}$
- D. 84 g mol$^{-1}$
Explanation: 40 + 12 + 48 = 100.
Particles
5. How many molecules are present in 0.5 mol of water?
- A. $3.011\times10^{23}$ (Correct)
- B. $6.022\times10^{23}$
- C. $1.204\times10^{24}$
- D. $0.5\times10^{22}$
Explanation: Number of molecules = 0.5 × Avogadro constant.
Gas volume
6. At STP, 2.24 L of an ideal gas corresponds to:
- A. 0.1 mol (Correct)
- B. 0.01 mol
- C. 1 mol
- D. 2 mol
Explanation: 22.4 L at STP corresponds to 1 mol, so 2.24 L corresponds to 0.1 mol.
Composition
7. Mass percent of hydrogen in water is closest to:
- A. 11.1% (Correct)
- B. 22.2%
- C. 5.6%
- D. 88.9%
Explanation: Hydrogen contributes 2 g in 18 g of water.
Empirical formula
8. An empirical formula represents:
- A. Simplest whole-number ratio of atoms (Correct)
- B. Exact molecular mass
- C. Structural arrangement
- D. Mass in grams
Explanation: Empirical formula gives the simplest atom ratio.
Stoichiometry basics
9. The coefficient in a balanced equation indicates:
- A. Mole ratio (Correct)
- B. Atomic radius
- C. Density ratio
- D. Bond order
Explanation: Balanced coefficients compare reacting and formed moles.
Stoichiometry basics
10. From $2H_2 + O_2 \to 2H_2O$, 1 mol of $O_2$ gives:
- A. 2 mol $H_2O$ (Correct)
- B. 1 mol $H_2O$
- C. 0.5 mol $H_2O$
- D. 4 mol $H_2O$
Explanation: The coefficient ratio of $O_2:H_2O$ is 1:2.
Mass relation
11. Number of moles in 4.4 g of $CO_2$ is:
- A. 0.1 (Correct)
- B. 0.2
- C. 0.05
- D. 1
Explanation: 4.4/44 = 0.1 mol.
Atoms vs molecules
12. One mole of helium gas contains:
- A. One mole of atoms (Correct)
- B. Two moles of atoms
- C. Half mole of atoms
- D. One mole of molecules each with two atoms
Explanation: Helium is monatomic.
Atoms vs molecules
13. One mole of $O_2$ contains:
- A. 2 mol oxygen atoms (Correct)
- B. 1 mol oxygen atoms
- C. 0.5 mol oxygen atoms
- D. 4 mol oxygen atoms
Explanation: Each $O_2$ molecule has 2 oxygen atoms.
Formula mass
14. Formula mass is generally used for:
- A. Ionic solids (Correct)
- B. Only gases
- C. Only covalent liquids
- D. Organic molecules only
Explanation: Ionic compounds are represented by formula units rather than discrete molecules.
Gram-atom
15. One gram-atom of sodium means:
- A. 23 g sodium (Correct)
- B. 1 g sodium
- C. 6.022 g sodium
- D. 46 g sodium
Explanation: Gram-atom means atomic mass expressed in grams.
Concentration
16. Molarity is defined as:
- A. Moles of solute per litre of solution (Correct)
- B. Moles of solute per kg solvent
- C. Gram of solute per litre solvent
- D. Equivalent per litre
Explanation: That is the definition of molarity.
Concentration
17. Molality depends on:
- A. Mass of solvent (Correct)
- B. Volume of solution
- C. Temperature directly
- D. Pressure directly
Explanation: Molality = moles of solute per kg of solvent.
Strength of solution
18. A 1 M NaCl solution contains:
- A. 1 mol NaCl in 1 L solution (Correct)
- B. 1 mol NaCl in 1 kg solvent
- C. 58.5 g NaCl in 1 kg solution
- D. 1 equivalent in 1 mL
Explanation: Molarity is based on litre of final solution.
Equivalent idea
19. Equivalent mass of an acid is its molar mass divided by:
- A. Basicity (Correct)
- B. Acidity of medium
- C. Valency of metal
- D. Number of atoms
Explanation: Equivalent mass for acids = molar mass / replaceable H$^+$ count.
Equivalent idea
20. Equivalent mass of $H_2SO_4$ in acid-base neutralisation is:
- A. 49 (Correct)
- B. 98
- C. 24.5
- D. 196
Explanation: Basicity of sulfuric acid is 2, so 98/2 = 49.
Mole conversion
21. Number of molecules in 18 g water is:
- A. $6.022\times10^{23}$ (Correct)
- B. $3.011\times10^{23}$
- C. $12.044\times10^{23}$
- D. $18\times10^{23}$
Explanation: 18 g of water is 1 mol.
NEET trap
22. Molar mass of $Na_2CO_3$ is:
- A. 106 (Correct)
- B. 84
- C. 62
- D. 53
Explanation: 46 from sodium, 12 from carbon, 48 from oxygen.
NEET trap
23. Mass of 0.25 mol NaOH is:
- A. 10 g (Correct)
- B. 20 g
- C. 40 g
- D. 5 g
Explanation: Molar mass of NaOH is 40, so mass = 0.25 × 40.
Volume relation
24. At STP, 11.2 L methane corresponds to:
- A. 0.5 mol (Correct)
- B. 1 mol
- C. 2 mol
- D. 0.25 mol
Explanation: 11.2 L is half of 22.4 L.
Chapter logic
25. The most useful starting step in a stoichiometry question is:
- A. Convert the given quantity into moles (Correct)
- B. Memorise density values
- C. Assume all reactants are limiting
- D. Convert everything into molarity
Explanation: Mole conversion is the core bridge between mass, particles, and volume.
Limiting reagent
26. The reactant consumed first in a reaction is called:
- A. Limiting reagent (Correct)
- B. Excess reagent
- C. Catalyst
- D. Indicator
Explanation: It limits the amount of product formed.
Limiting reagent
27. In $2H_2 + O_2 \to 2H_2O$, if 1 mol $H_2$ reacts with 1 mol $O_2$, the limiting reagent is:
- A. $H_2$ (Correct)
- B. $O_2$
- C. Both equal
- D. Cannot be decided
Explanation: 1 mol $H_2$ needs only 0.5 mol $O_2$, so hydrogen gets consumed first.
Limiting reagent
28. If 4 mol NH$_3$ are formed theoretically but only 3 mol are actually obtained, percent yield is:
- A. 75% (Correct)
- B. 80%
- C. 66.7%
- D. 25%
Explanation: Percent yield = actual/theoretical × 100.
Yield
29. Theoretical yield is calculated from:
- A. Limiting reagent (Correct)
- B. Excess reagent only
- C. Catalyst amount
- D. Pressure
Explanation: Maximum possible product is based on the limiting reagent.
Excess reagent
30. In a reaction mixture, the reagent left after completion is:
- A. Excess reagent (Correct)
- B. Limiting reagent
- C. Product
- D. Intermediate
Explanation: It remains unused after the limiting reagent is exhausted.
Stoichiometry
31. For $N_2 + 3H_2 \to 2NH_3$, 3 mol $H_2$ produce:
- A. 2 mol $NH_3$ (Correct)
- B. 1 mol $NH_3$
- C. 3 mol $NH_3$
- D. 6 mol $NH_3$
Explanation: The coefficient ratio is 3:2.
Stoichiometry
32. For $N_2 + 3H_2 \rightarrow 2NH_3$, 28 g $N_2$ with 3 g $H_2$ gives maximum ammonia equal to:
- A. 34 g
- B. 17 g (Correct)
- C. 28 g
- D. 14 g
Explanation: 28 g $N_2$ = 1 mol and 3 g $H_2$ = 1.5 mol, so $H_2$ is limiting and forms 1 mol NH$_3$ = 17 g? Wait check. 3 mol H2 gives 2 mol NH3, so 1.5 mol gives 1 mol NH3 = 17 g.
Stoichiometry
33. For $N_2 + 3H_2 \to 2NH_3$, if 28 g $N_2$ and 3 g $H_2$ react completely, mass of NH$_3$ formed is:
- A. 17 g (Correct)
- B. 34 g
- C. 14 g
- D. 28 g
Explanation: 3 g hydrogen is 1.5 mol, so it is limiting and produces 1 mol ammonia, i.e. 17 g.
Purity
34. A 10 g sample of limestone containing 80% $CaCO_3$ has pure $CaCO_3$ mass:
- A. 8 g (Correct)
- B. 2 g
- C. 10 g
- D. 12.5 g
Explanation: Pure substance mass = 80% of 10 g.
Purity
35. Moles of $CaCO_3$ in that pure sample are:
- A. 0.08 (Correct)
- B. 0.8
- C. 0.04
- D. 8
Explanation: Molar mass is 100, so moles = 8/100.
Product mass
36. On decomposition, 1 mol $CaCO_3$ forms 1 mol of:
- A. $CO_2$ (Correct)
- B. $O_2$
- C. $Ca(OH)_2$
- D. $H_2$
Explanation: $CaCO_3 \to CaO + CO_2$.
Product mass
37. Mass of $CO_2$ from complete decomposition of 10 g pure $CaCO_3$ is:
- A. 4.4 g (Correct)
- B. 2.2 g
- C. 5.6 g
- D. 10 g
Explanation: 10 g $CaCO_3$ is 0.1 mol and gives 0.1 mol $CO_2$ = 4.4 g.
Molarity
38. If 0.5 mol solute is present in 250 mL solution, molarity is:
- A. 2 M (Correct)
- B. 0.5 M
- C. 1 M
- D. 4 M
Explanation: 0.5 mol / 0.25 L = 2 M.
Molality
39. If 0.2 mol solute is dissolved in 500 g water, molality is:
- A. 0.4 m (Correct)
- B. 0.2 m
- C. 2.5 m
- D. 4 m
Explanation: 0.2 mol per 0.5 kg solvent gives 0.4 m.
Normality
40. Normality of 1 M $H_2SO_4$ in acid-base reactions is:
- A. 2 N (Correct)
- B. 1 N
- C. 0.5 N
- D. 4 N
Explanation: Normality = molarity × n-factor = 1 × 2.
Dilution
41. For dilution, the correct relation is:
- A. $M_1V_1=M_2V_2$ (Correct)
- B. $M_1+V_1=M_2+V_2$
- C. $M_1/V_1=M_2/V_2$
- D. $M_1M_2=V_1V_2$
Explanation: Moles of solute remain constant during dilution.
Dilution
42. Volume of water to be added to 100 mL of 2 M solution to make it 1 M is:
- A. 100 mL (Correct)
- B. 50 mL
- C. 200 mL
- D. 25 mL
Explanation: Final volume should become 200 mL, so add 100 mL water.
Equivalent concept
43. Equivalent weight of $KMnO_4$ in acidic medium is molar mass divided by:
- A. 5 (Correct)
- B. 2
- C. 1
- D. 8
Explanation: Mn changes from +7 to +2, so n-factor is 5.
Equivalent concept
44. Equivalent weight of $KMnO_4$ in neutral/basic medium is molar mass divided by:
- A. 3 (Correct)
- B. 5
- C. 2
- D. 1
Explanation: In basic or neutral medium Mn goes from +7 to +4.
Mixture logic
45. If two reactants are taken exactly in stoichiometric ratio, then:
- A. Neither remains in excess (Correct)
- B. Both become limiting separately
- C. Product is zero
- D. Catalyst is required
Explanation: Both are completely consumed together.
Mass percent
46. A solution with 10 g solute in 90 g water has mass percent of solute:
- A. 10% (Correct)
- B. 9%
- C. 11.1%
- D. 90%
Explanation: Mass percent is solute mass divided by solution mass times 100.
Atomicity
47. One mole of $P_4$ contains phosphorus atoms equal to:
- A. 4 mol atoms (Correct)
- B. 1 mol atoms
- C. 0.25 mol atoms
- D. 2 mol atoms
Explanation: Each molecule of $P_4$ contains four phosphorus atoms.
Water of crystallization
48. Molar mass of washing soda $Na_2CO_3\cdot10H_2O$ is:
- A. 286 (Correct)
- B. 106
- C. 180
- D. 304
Explanation: 106 + 10×18 = 286.
Practical stoichiometry
49. A 5 g impure sample containing 50% $CaCO_3$ liberates $CO_2$ on treatment with acid. Mass of $CO_2$ formed is:
- A. 1.1 g (Correct)
- B. 2.2 g
- C. 0.55 g
- D. 4.4 g
Explanation: Pure $CaCO_3$ is 2.5 g = 0.025 mol, hence $CO_2$ formed is 0.025 mol = 1.1 g.
Chapter logic
50. Percent yield can never exceed 100% in a correct pure-product calculation because:
- A. Actual yield cannot exceed theoretical maximum (Correct)
- B. Catalysts reduce yield
- C. Limiting reagent disappears
- D. Mass is never conserved
Explanation: Theoretical yield already represents the maximum possible product.
Atomic mass
51. Average atomic mass of chlorine is about 35.5 because chlorine exists as:
- A. A mixture of isotopes (Correct)
- B. A polymer
- C. A dimer only
- D. An ion pair
Explanation: Weighted average of isotopic masses gives average atomic mass.
Isotopes
52. Atoms of same element having different mass numbers are:
- A. Isotopes (Correct)
- B. Isobars
- C. Isoelectronic species
- D. Allotropes
Explanation: They have same atomic number but different neutrons.
Isobars
53. Atoms of different elements having same mass number are:
- A. Isobars (Correct)
- B. Isotopes
- C. Isotones
- D. Allotropes
Explanation: Their atomic numbers differ but mass number is same.
Average mass
54. If an element has isotopes of masses 10 and 11 in ratio 1:3, average atomic mass is:
- A. 10.75 (Correct)
- B. 10.5
- C. 11
- D. 10.25
Explanation: Weighted average = (10×1 + 11×3)/4.
Average mass
55. An element has 75% isotope of mass 24 and 25% isotope of mass 26. Average mass is:
- A. 24.5 (Correct)
- B. 25
- C. 24.75
- D. 25.5
Explanation: Weighted average = 24×0.75 + 26×0.25.
Empirical formula
56. If molecular formula is $C_6H_{12}O_6$, empirical formula is:
- A. $CH_2O$ (Correct)
- B. $C_6H_{12}O_6$
- C. $C_3H_6O_3$
- D. $CHO$
Explanation: Divide all subscripts by 6.
Empirical formula
57. A compound contains C = 40%, H = 6.67%, O = 53.33%. Its empirical formula is:
- A. $CH_2O$ (Correct)
- B. $C_2H_4O_2$
- C. $CHO$
- D. $CH_3O$
Explanation: Convert percentages to mole ratio: 40/12 : 6.67/1 : 53.33/16 ≈ 1 : 2 : 1.
Molecular formula
58. If the empirical formula is $CH_2O$ and molar mass is 180, molecular formula is:
- A. $C_6H_{12}O_6$ (Correct)
- B. $C_3H_6O_3$
- C. $CH_2O$
- D. $C_2H_4O_2$
Explanation: Empirical formula mass is 30, multiplier is 180/30 = 6.
Combustion analysis
59. A hydrocarbon on complete combustion gives 4.4 g $CO_2$ and 1.8 g water. Mass of carbon in the sample is:
- A. 1.2 g (Correct)
- B. 2.4 g
- C. 0.6 g
- D. 4.4 g
Explanation: 4.4 g $CO_2$ corresponds to 0.1 mol carbon = 1.2 g.
Combustion analysis
60. From the same data, mass of hydrogen in the sample is:
- A. 0.2 g (Correct)
- B. 0.4 g
- C. 1.8 g
- D. 0.1 g
Explanation: 1.8 g water is 0.1 mol water, so hydrogen atoms are 0.2 mol = 0.2 g.
Oxidation state
61. Oxidation number of oxygen in most compounds is:
- A. -2 (Correct)
- B. -1
- C. 0
- D. +2
Explanation: This is the usual oxidation state except in peroxides, superoxides, and OF$_2$.
Oxidation state
62. Oxidation number of hydrogen in most compounds is:
- A. +1 (Correct)
- B. -1
- C. 0
- D. +2
Explanation: Hydrogen is usually +1 except in metal hydrides.
Oxidation state
63. Oxidation state of sulfur in $H_2SO_4$ is:
- A. +6 (Correct)
- B. +4
- C. +2
- D. 0
Explanation: Let sulfur be x: 2(+1) + x + 4(-2) = 0, so x = +6.
Oxidation state
64. Oxidation state of Mn in $KMnO_4$ is:
- A. +7 (Correct)
- B. +2
- C. +4
- D. +6
Explanation: K is +1 and oxygen contributes -8, so Mn must be +7.
Oxidation state
65. Oxidation state of Cr in $K_2Cr_2O_7$ is:
- A. +6 (Correct)
- B. +3
- C. +7
- D. +2
Explanation: 2(+1) + 2x + 7(-2) = 0 gives x = +6.
Redox basics
66. Oxidation is defined as:
- A. Loss of electrons (Correct)
- B. Gain of electrons
- C. Gain of neutrons
- D. Loss of protons
Explanation: LEO: loss of electrons is oxidation.
Redox basics
67. Reduction is defined as:
- A. Gain of electrons (Correct)
- B. Loss of electrons
- C. Gain of mass
- D. Loss of neutrons
Explanation: GER: gain of electrons is reduction.
Redox agent
68. The species that gets reduced acts as:
- A. Oxidising agent (Correct)
- B. Reducing agent
- C. Catalyst
- D. Solvent
Explanation: It oxidises the other species while itself undergoing reduction.
Redox agent
69. The species that gets oxidised acts as:
- A. Reducing agent (Correct)
- B. Oxidising agent
- C. Indicator
- D. Buffer
Explanation: It donates electrons to reduce the other species.
n-factor
70. n-factor of $H_3PO_4$ in complete neutralisation is:
- A. 3 (Correct)
- B. 1
- C. 2
- D. 4
Explanation: All three acidic hydrogens can be replaced in complete neutralisation.
n-factor
71. n-factor of NaOH in acid-base reaction is:
- A. 1 (Correct)
- B. 2
- C. 3
- D. 0.5
Explanation: One mole NaOH furnishes one mole OH$^-$.
Equivalent concept
72. Equivalent mass of $Na_2CO_3$ in acid-base reaction is:
- A. 53 (Correct)
- B. 106
- C. 26.5
- D. 84
Explanation: It accepts 2 H$^+$, so equivalent mass = 106/2.
Isotopic abundance
73. Average atomic mass is best described as:
- A. Weighted mean of isotopic masses (Correct)
- B. Mass of most abundant isotope only
- C. Mass number of the lightest isotope
- D. Sum of atomic and neutron numbers
Explanation: Isotopic abundance determines the weighting.
Formula determination
74. A compound with empirical formula mass 30 and molecular mass 60 has molecular formula factor:
- A. 2 (Correct)
- B. 1
- C. 3
- D. 0.5
Explanation: Factor = molecular mass / empirical formula mass.
Chapter logic
75. In formula problems, the best sequence is:
- A. Convert percent to moles, reduce to simplest ratio, then scale using molar mass (Correct)
- B. Start with oxidation number
- C. Use molarity first
- D. Assume empirical and molecular formula are same
Explanation: That is the standard reliable method.
Gaseous laws
76. At constant temperature, pressure of a fixed mass of gas is inversely proportional to volume according to:
- A. Boyle's law (Correct)
- B. Charles' law
- C. Avogadro's law
- D. Graham's law
Explanation: Boyle's law applies for isothermal changes.
Gaseous laws
77. At constant pressure, volume is directly proportional to absolute temperature according to:
- A. Charles' law (Correct)
- B. Boyle's law
- C. Dalton's law
- D. Henry's law
Explanation: Charles' law uses temperature on Kelvin scale.
Gaseous laws
78. At constant volume, pressure is directly proportional to absolute temperature according to:
- A. Gay-Lussac law (Correct)
- B. Boyle's law
- C. Raoult law
- D. Ostwald law
Explanation: Pressure varies directly with Kelvin temperature at constant volume.
Ideal gas
79. The ideal gas equation is:
- A. $PV=nRT$ (Correct)
- B. $P=nVRT$
- C. $PV=RT$
- D. $P/T=nR/V^2$
Explanation: This combines major simple gas laws.
Gas constant
80. Value of R in L atm mol$^{-1}$ K$^{-1}$ is approximately:
- A. 0.0821 (Correct)
- B. 8.314
- C. 22.4
- D. 1.987
Explanation: This is the commonly used value in these units.
Density
81. Density of a gas is related to molar mass by:
- A. $d=\frac{PM}{RT}$ (Correct)
- B. $d=\frac{RT}{PM}$
- C. $d=\frac{P}{MRT}$
- D. $d=PMRT$
Explanation: From $PV=nRT$ and $n=m/M$, density becomes $m/V = PM/RT$.
Vapour density
82. Vapour density is equal to:
- A. Molar mass/2 (Correct)
- B. Molar mass
- C. 2×molar mass
- D. Density at STP only
Explanation: On the hydrogen scale, vapour density is half the molar mass.
Mixture of gases
83. Total pressure of a mixture of non-reacting gases is:
- A. Sum of partial pressures (Correct)
- B. Product of pressures
- C. Average of partial pressures
- D. Always 1 atm
Explanation: This is Dalton's law.
Partial pressure
84. If mole fraction of nitrogen in air sample is 0.8 and total pressure is 2 atm, partial pressure of nitrogen is:
- A. 1.6 atm (Correct)
- B. 0.4 atm
- C. 2.5 atm
- D. 0.8 atm
Explanation: Partial pressure = mole fraction × total pressure.
Mole fraction
85. Mole fraction is a unitless ratio of:
- A. Moles of a component to total moles (Correct)
- B. Mass of solute to solvent mass
- C. Volume to temperature
- D. Equivalent to molarity
Explanation: It measures composition independent of units.
Kinetic theory
86. Average kinetic energy of an ideal gas depends only on:
- A. Temperature (Correct)
- B. Pressure
- C. Volume
- D. Nature of gas
Explanation: Average KE is proportional to absolute temperature.
Kinetic theory
87. RMS speed of gas molecules is given by:
- A. $\sqrt{3RT/M}$ (Correct)
- B. $\sqrt{RT/3M}$
- C. $3RT/M$
- D. $\sqrt{M/3RT}$
Explanation: This is the standard rms velocity formula.
Graham law
88. According to Graham's law, rate of diffusion is inversely proportional to:
- A. Square root of molar mass (Correct)
- B. Molar mass
- C. Density directly
- D. Pressure directly
Explanation: Lighter gases diffuse faster.
Graham law
89. If hydrogen diffuses four times faster than a gas X, molar mass of X is:
- A. 32 (Correct)
- B. 16
- C. 8
- D. 64
Explanation: Rate ratio = sqrt(Mx/2) = 4, so Mx = 32.
Real gases
90. A real gas approaches ideal behavior most closely at:
- A. Low pressure and high temperature (Correct)
- B. High pressure and low temperature
- C. High pressure and high temperature
- D. Low pressure and low temperature
Explanation: Intermolecular attractions become least important there.
Compression factor
91. For an ideal gas, compressibility factor Z equals:
- A. 1 (Correct)
- B. 0
- C. Depends on pressure only
- D. Greater than 1 always
Explanation: For ideal gases, PV = nRT exactly.
Stoichiometric gas
92. At the same temperature and pressure, equal volumes of gases contain:
- A. Equal number of molecules (Correct)
- B. Equal masses
- C. Equal densities
- D. Equal molecular speed
Explanation: This is Avogadro's law.
STP use
93. At STP, one mole of an ideal gas occupies:
- A. 22.4 L (Correct)
- B. 24.4 L
- C. 11.2 L
- D. 44.8 L
Explanation: That is the standard textbook value used in NEET problems.
Temperature scale
94. Absolute zero on Celsius scale is:
- A. -273.15$^\circ$C (Correct)
- B. 0$^\circ$C
- C. 273$^\circ$C
- D. -100$^\circ$C
Explanation: Kelvin = Celsius + 273.15.
Gas mixture
95. Mole fraction of a component in an ideal gas mixture is numerically equal to:
- A. Its partial pressure divided by total pressure (Correct)
- B. Its molarity
- C. Its mass fraction always
- D. Its vapour density
Explanation: For ideal mixtures, $x_i = p_i/P$.
Equation use
96. For 2 mol ideal gas at 300 K and 0.0821 L atm mol$^{-1}$ K$^{-1}$, $PV$ equals:
- A. 49.26 L atm (Correct)
- B. 24.63 L atm
- C. 82.1 L atm
- D. 600 L atm
Explanation: $PV=nRT=2\times0.0821\times300$.
State variables
97. Which is an intensive property?
- A. Pressure (Correct)
- B. Volume
- C. Mass
- D. Number of moles
Explanation: Intensive properties do not depend on system size.
State variables
98. Which is an extensive property?
- A. Volume (Correct)
- B. Temperature
- C. Pressure
- D. Density
Explanation: Extensive properties scale with amount of matter.
Chapter logic
99. The main reason ideal gas law is so useful in mole concept is that it connects:
- A. Moles with measurable P, V, and T (Correct)
- B. Oxidation number with structure
- C. Mass percent with isotopes
- D. Equivalent weight with catalysis
Explanation: It lets us convert gas data directly into amount of substance.
NEET trap
100. When temperature is doubled from 300 K to 600 K at constant pressure, gas volume becomes:
- A. Double (Correct)
- B. Half
- C. Four times
- D. Unchanged
Explanation: Charles' law uses Kelvin temperature, so doubling K doubles V.
Redox balance
101. In acidic medium, $MnO_4^-$ is commonly reduced to:
- A. $Mn^{2+}$ (Correct)
- B. $MnO_2$
- C. $Mn$
- D. $MnO_4^{2-}$
Explanation: Permanganate reduces to Mn$^{2+}$ in acidic medium.
Redox balance
102. In acidic medium, dichromate ion is reduced to:
- A. $Cr^{3+}$ (Correct)
- B. $CrO_4^{2-}$
- C. Cr metal
- D. $Cr^{6+}$
Explanation: Dichromate is a strong oxidising agent in acid medium.
Oxidation number method
103. The first step in oxidation number method is usually:
- A. Assign oxidation numbers (Correct)
- B. Write molarity
- C. Calculate density
- D. Find boiling point
Explanation: This reveals oxidation and reduction changes.
Oxidation number method
104. The sum of increase in oxidation number must equal:
- A. Sum of decrease in oxidation number (Correct)
- B. Total pressure
- C. Number of atoms always
- D. Number of electrons in atom
Explanation: Electron loss and gain must balance.
Ionic equation
105. Spectator ions are those that:
- A. Remain unchanged during ionic reaction (Correct)
- B. Act as catalysts only
- C. Always form precipitate
- D. Always get oxidised
Explanation: They do not participate in the net ionic change.
Equivalent relation
106. In titration, equivalents of acid at end point are equal to:
- A. Equivalents of base (Correct)
- B. Moles of solvent
- C. Total mass of solution
- D. Partial pressures
Explanation: Neutralisation follows equivalent equality at equivalence point.
Titration
107. If 25 mL of 0.1 N HCl neutralises NaOH solution, the NaOH solution containing 20 mL sample has normality:
- A. 0.125 N (Correct)
- B. 0.1 N
- C. 0.08 N
- D. 0.2 N
Explanation: $N_1V_1=N_2V_2$, so $0.1\times25=N\times20$.
n-factor
108. n-factor of oxalic acid in redox titration is:
- A. 2 (Correct)
- B. 1
- C. 5
- D. 4
Explanation: Each oxalate ion loses two electrons on oxidation to carbon dioxide.
n-factor
109. n-factor of $K_2Cr_2O_7$ in acidic medium is:
- A. 6 (Correct)
- B. 3
- C. 2
- D. 1
Explanation: Each dichromate ion gains 6 electrons in acid medium.
Balancing
110. A balanced redox equation must satisfy conservation of:
- A. Mass and charge (Correct)
- B. Mass only
- C. Charge only
- D. Volume only
Explanation: Both atoms and net charge must balance.
Oxidation number
111. Oxidation state of carbon in $CO$ is:
- A. +2 (Correct)
- B. +4
- C. 0
- D. -2
Explanation: Oxygen is -2, so carbon must be +2.
Oxidation number
112. Oxidation state of carbon in $CO_2$ is:
- A. +4 (Correct)
- B. +2
- C. 0
- D. -4
Explanation: Two oxygens contribute -4, so carbon is +4.
Disproportionation
113. A reaction in which the same species is simultaneously oxidised and reduced is:
- A. Disproportionation (Correct)
- B. Neutralisation
- C. Hydrolysis
- D. Addition
Explanation: The same element undergoes both changes in oxidation state.
Comproportionation
114. When two oxidation states of the same element combine to form an intermediate oxidation state, the reaction is:
- A. Comproportionation (Correct)
- B. Displacement
- C. Condensation
- D. Polymerisation
Explanation: It is the reverse tendency of disproportionation.
Stoichiometric redox
115. In acidic medium, one mole $KMnO_4$ corresponds to how many equivalents?
- A. 5 (Correct)
- B. 1
- C. 2
- D. 3
Explanation: Equivalent count equals n-factor in redox.
Stoichiometric redox
116. A 1 M $KMnO_4$ solution in acidic medium is:
- A. 5 N (Correct)
- B. 1 N
- C. 0.2 N
- D. 2 N
Explanation: Normality = molarity × n-factor = 1 × 5.
Acid-base vs redox
117. Normality depends on reaction because:
- A. n-factor depends on the process involved (Correct)
- B. Molar mass changes with temperature
- C. Volume is always fixed
- D. Atoms are created
Explanation: The same substance can have different n-factor in different reactions.
Data logic
118. If 10 mL of 0.2 N acid neutralises a base sample, the sample contains base equivalents equal to:
- A. 0.002 (Correct)
- B. 2
- C. 0.02
- D. 0.0002
Explanation: Equivalents = N × V in litres = 0.2 × 0.01.
Error trap
119. Which quantity changes with temperature and therefore is avoided in many precise stoichiometric calculations?
- A. Molarity (Correct)
- B. Molality
- C. Mole fraction
- D. Mass percent
Explanation: Molarity depends on solution volume, which changes with temperature.
Error trap
120. Molality is preferred in colligative calculations because it:
- A. Does not depend on temperature (Correct)
- B. Is based on litre of solution
- C. Uses equivalents
- D. Is always greater than molarity
Explanation: Mass of solvent stays unchanged with temperature.
Mass-volume link
121. A gas law based mole calculation is most reliable when the gas behaves:
- A. Nearly ideally (Correct)
- B. As a solid
- C. With strong association always
- D. At very high pressure only
Explanation: Ideal behaviour makes $PV=nRT$ accurate.
Applied stoichiometry
122. A sample of impure NaOH weighing 8 g is 50% pure. Number of moles of pure NaOH present is:
- A. 0.1 (Correct)
- B. 0.2
- C. 0.4
- D. 0.05
Explanation: Pure NaOH mass is 4 g and molar mass is 40.
Applied stoichiometry
123. That much pure NaOH would neutralise moles of HCl equal to:
- A. 0.1 (Correct)
- B. 0.2
- C. 0.05
- D. 0.01
Explanation: NaOH and HCl react in 1:1 mole ratio.
Applied logic
124. The cleanest way to avoid stoichiometric mistakes is to track:
- A. Units at every step (Correct)
- B. Only final answer
- C. Only formula names
- D. Only atomic numbers
Explanation: Unit tracking catches many mole and volume mistakes.
Chapter integration
125. The chapter 'Some Basic Concepts of Chemistry' is fundamentally about:
- A. Relating mass, particles, solution strength, and reaction ratios through mole language (Correct)
- B. Only naming compounds
- C. Only drawing atomic models
- D. Only balancing redox equations
Explanation: Mole language unifies all these quantitative ideas.

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