Some basic concept of chemistry
Class 11 Chemistry: Chapter 1 – Some Basic Concepts of Chemistry (20 MCQs)
Q.1. What is the SI unit of amount of substance? (a) Kilogram (b) Mole (c) Pascal (d) Kelvin
Answer: (b) Mole Detailed Answer: The International System of Units (SI) has seven base units. The unit for the amount of substance is the mole (mol). Kilogram (kg) is the unit for mass, Pascal (Pa) for pressure, and Kelvin (K) for thermodynamic temperature.
Q.2. How many significant figures are present in the measurement 0.00250? (a) Five (b) Four (c) Three (d) Two
Answer: (c) Three Detailed Answer: The rules for significant figures are:
- All non-zero digits are significant.
- Zeros preceding the first non-zero digit (leading zeros) are not significant. (Here, the first two zeros after the decimal are not significant).
- Zeros between two non-zero digits (trapped zeros) are significant. (Not applicable here).
- Trailing zeros (zeros at the end of a number) are significant if the number contains a decimal point. (The final zero is significant). Therefore, the significant figures are 2, 5, and 0, making a total of three significant figures.
Q.3. Which law of chemical combination is best illustrated by the formation of Carbon Monoxide (CO) and Carbon Dioxide (CO2)? (a) Law of Conservation of Mass (b) Law of Constant Proportions (c) Law of Multiple Proportions (d) Gay-Lussac’s Law of Gaseous Volumes
Answer: (c) Law of Multiple Proportions Detailed Answer: The Law of Multiple Proportions states that if two elements can combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element are in a ratio of small whole numbers.
- In CO and CO2, a fixed mass of Carbon (12 parts) combines with different masses of Oxygen:
- CO: 12 parts C:16 parts O
- CO2: 12 parts C:32 parts O
- The ratio of masses of oxygen (16:32) is 1:2, which is a simple whole-number ratio.
Q.4. What is the mass of one molecule of glucose (C6H12O6) in grams? (Atomic masses: C=12 u, H=1 u, O=16 u; NA=6.022×1023) (a) 180 g (b) 3.0×10−22 g (c) 1.66×10−24 g (d) 2.99×10−23 g
Answer: (d) 2.99×10−23 g Detailed Answer:
- Molar Mass of Glucose (C6H12O6): M=(6×12)+(12×1)+(6×16)=72+12+96=180 g/mol
- Mass of One Molecule: The mass of one mole (6.022×1023 molecules) is 180 g. Mass of one molecule=Avogadro’s Number(NA)Molar MassMass=6.022×1023 molecules/mol180 g≈29.89×10−23 g/molecule≈2.99×10−22 g
Q.5. Which of the following is an example of a homogeneous mixture? (a) Sand and water (b) Sugar solution (c) Oil and water (d) Copper fillings and sulphur powder
Answer: (b) Sugar solution Detailed Answer: A homogeneous mixture is a mixture in which the components mix uniformly, and the composition is the same throughout the mixture. A sugar solution (sugar dissolved in water) is an example of a homogeneous mixture. The other options are examples of heterogeneous mixtures, where the components remain separate.
Q.6. The empirical formula of a compound is CH2O, and its molecular mass is 180 g. What is its molecular formula? (a) CH2O (b) C6H12O6 (c) C2H4O2 (d) C3H6O3
Answer: (b) C6H12O6 Detailed Answer:
- Empirical Formula Mass (CH2O): EFM=(1×12)+(2×1)+(1×16)=30 u
- Calculate ‘n’ factor: n=Empirical Formula MassMolecular Mass=30 g180 g=6
- Molecular Formula: Molecular Formula=n×(Empirical Formula)=6×(CH2O)=C6H12O6
Q.7. The maximum number of molecules is present in: (a) 15 L of H2 gas at STP (b) 5 L of N2 gas at STP (c) 0.5 g of H2 gas (d) 10 g of O2 gas
Answer: (a) 15 L of H2 gas at STP Detailed Answer: The number of molecules is directly proportional to the number of moles.
- For gases at STP: The number of moles is directly proportional to volume (Avogadro’s Law).
- (a) 15 L of H2: Moles ∝15
- (b) 5 L of N2: Moles ∝5
- For the given masses (comparing the moles):
- (c) 0.5 g of H2 (Molar Mass=2 g/mol): Moles=2 g/mol0.5 g=0.25 mol
- (d) 10 g of O2 (Molar Mass=32 g/mol): Moles=32 g/mol10 g≈0.3125 mol
- The volume of 1 mole of gas at STP is 22.7 L.
- (a) 15 L of H2: Moles=22.7 L/mol15 L≈0.66 mol
- Comparing the number of moles: 0.66>0.3125>0.25>moles of 5 L gas. Option (a) has the maximum number of moles, and hence the maximum number of molecules.
Q.8. Molarity is defined as: (a) Moles of solute per kg of solvent. (b) Moles of solute per L of solution. (c) Grams of solute per L of solution. (d) Grams of solute per kg of solvent.
Answer: (b) Moles of solute per L of solution. Detailed Answer: Molarity (M) is a measure of the concentration of a solute in a solution, defined as the number of moles of solute dissolved per litre of the solution.
Molarity(M)=Volume of solution in litresMoles of solute
Q.9. Which of the following concentration terms is independent of temperature? (a) Molarity (b) Normality (c) Molality (d) Volume percentage
Answer: (c) Molality Detailed Answer:
- Molarity (M), Normality (N), and Volume Percentage are all dependent on the volume of the solution. Since volume changes with temperature, these concentration terms also change with temperature.
- Molality (m) is defined as the moles of solute per mass of solvent (in kg). Mass does not change with temperature, so molality is independent of temperature.
Q.10. 4 L of N2 gas reacts with 9 L of H2 gas to form ammonia (NH3). What is the maximum volume of NH3 that can be formed? (Assume the reaction occurs at constant temperature and pressure). The balanced reaction is: N2(g)+3H2(g)→2NH3(g) (a) 4 L (b) 6 L (c) 8 L (d) 12 L
Answer: (b) 6 L Detailed Answer: The problem involves Gay-Lussac’s Law of Gaseous Volumes and the concept of Limiting Reagent. The volume ratios are the same as the mole ratios.
Ratio (Volume):Initial Volume:N2(g)14 L+:3H2(g)39 L→:2NH3(g)20 L
- Find the Limiting Reagent (LR):
- N2 required for 9 L of H2: 9 L H2×3 L H21 L N2=3 L N2. Since we have 4 L of N2 (which is more than 3 L), N2 is in excess.
- H2 required for 4 L of N2: 4 L N2×1 L N23 L H2=12 L H2. Since we only have 9 L of H2, H2 is the Limiting Reagent.
- Calculate the Volume of NH3 formed using the LR (H2): 9 L H2×3 L H22 L NH3=6 L NH3 The maximum volume of NH3 formed is 6 L.
Q.11. How many atoms are present in 46 g of sodium (Na)? (Atomic mass of Na=23 u; NA=6.022×1023 atoms/mol) (a) 6.022×1023 (b) 1.2044×1024 (c) 2.4088×1024 (d) 3.011×1023
Answer: (b) 1.2044×1024 Detailed Answer:
- Calculate Moles of Na: Moles=Atomic MassGiven Mass=23 g/mol46 g=2 moles
- Calculate Number of Atoms: Number of Atoms=Moles×NANumber of Atoms=2×(6.022×1023)=12.044×1023=1.2044×1024 atoms
Q.12. The temperature of a substance is 25∘C. What is this temperature on the Kelvin scale? (a) 298.15 K (b) 25 K (c) 248.15 K (d) 273.15 K
Answer: (a) 298.15 K Detailed Answer: The relationship between the Celsius scale (∘C) and the Kelvin scale (K) is:
Temperature in K=Temperature in ∘C+273.15
Temperature in K=25+273.15=298.15 K
Q.13. What is the mass percent of oxygen in water (H2O)? (Atomic masses: H=1.0 u, O=16.0 u) (a) 88.89% (b) 11.11% (c) 66.66% (d) 33.33%
Answer: (a) 88.89% Detailed Answer:
- Molar Mass of H2O: M=(2×1.0)+(1×16.0)=18.0 g/mol
- Mass of Oxygen in one mole of H2O: 16.0 g
- Mass Percent of Oxygen: Mass % of O=Molar Mass of H2OMass of Oxygen×100Mass % of O=18.016.0×100≈0.8888×100=88.89%
Q.14. Which of the following is a unit of density? (a) kg⋅m (b) g⋅L (c) g/cm3 (d) mol/L
Answer: (c) g/cm3 Detailed Answer: Density is defined as mass per unit volume.
Density=VolumeMass
- (a) kg⋅m is a unit of momentum or impulse.
- (c) g/cm3 (gram per cubic centimetre) is a valid unit of mass/volume (density).
- (d) mol/L (moles per litre) is a unit of molarity (concentration).
Q.15. When 2.0 L of a 5 M solution is diluted to 10.0 L, what is the final molarity of the solution? (a) 1.0 M (b) 0.5 M (c) 2.5 M (d) 0.1 M
Answer: (a) 1.0 M Detailed Answer: This is a dilution problem, which follows the formula:
M1V1=M2V2
Where:
- M1=5 M (Initial Molarity)
- V1=2.0 L (Initial Volume)
- M2=? (Final Molarity)
- V2=10.0 L (Final Volume) M2=V2M1V1=10.0 L5 M×2.0 L=1010=1.0 M
Q.16. Which statement about the Law of Conservation of Mass is correct? (a) Mass can be created but not destroyed. (b) The total mass of reactants is less than the total mass of products. (c) Matter can neither be created nor destroyed. (d) The mass of the system remains unchanged in a physical change only.
Answer: (c) Matter can neither be created nor destroyed. Detailed Answer: The Law of Conservation of Mass (or Matter) states that in a closed system, the mass of the system remains constant, meaning matter can neither be created nor destroyed by any physical or chemical change. In a chemical reaction, the total mass of the reactants must equal the total mass of the products.
Q.17. How many moles of methane (CH4) are required to produce 22 g of CO2 upon complete combustion? (Atomic masses: C=12 u, O=16 u) (a) 1.0 mol (b) 0.5 mol (c) 2.0 mol (d) 0.25 mol
Answer: (b) 0.5 mol Detailed Answer:
- Balanced Reaction: CH4(g)+2O2(g)→CO2(g)+2H2O(l)
- Molar Mass of CO2: M=12+(2×16)=44 g/mol
- Moles of CO2 produced: Moles=44 g/mol22 g=0.5 mol
- Stoichiometry (Mole Ratio): From the balanced equation, 1 mol of CH4 produces 1 mol of CO2.
- Moles of CH4 required: To produce 0.5 mol of CO2, 0.5 mol of CH4 is required.
Q.18. The number of significant figures in Avogadro’s number (6.022×1023) is: (a) 26 (b) 23 (c) 4 (d) 1
Answer: (c) 4 Detailed Answer: In scientific notation, all digits in the mantissa (the part before the power of 10) are considered significant figures.
- In 6.022×1023, the mantissa is 6.022.
- This number has four significant figures (6, 0, 2, and 2). The exponential part (1023) does not affect the number of significant figures.
Q.19. 1 amu (atomic mass unit) is equal to:
(a) 161 of the mass of an oxygen atom
(b) The mass of a proton
(c) 121 of the mass of a C-12 isotope atom
(d) The mass of a hydrogen atom
Answer: (c) 121 of the mass of a C-12 isotope atom Detailed Answer: The atomic mass unit (amu) or unified mass (u) is defined as exactly one-twelfth (1/12) of the mass of an atom of carbon-12 isotope.
Q.20. Which of the following prefixes corresponds to 10−9? (a) Mega (b) Giga (c) Nano (d) Micro
Answer: (c) Nano Detailed Answer:
- Nano (n) is the prefix for 10−9.
- Mega (M) is the prefix for 106.
- Giga (G) is the prefix for 109.
- Micro (μ) is the prefix for 10−6.