Some Basic Concepts of Chemistry MCQs (Mixed) with Answers
Some Basic Concepts of Chemistry is the chapter where you build the foundation for numericals: mole concept, molar mass, Avogadro’s number, empirical vs molecular formula, and basic stoichiometry. If you understand these, balancing equations and solving higher-level problems becomes easier. This mixed MCQ set helps you revise definitions and quick calculations in one place. Each question has the correct answer with a short explanation.
MCQs (Mixed Practice)
Question 1
Avogadro’s number represents the number of particles present in:
A. 1 g of any substance
B. 1 mole of any substance
C. 1 litre of any gas at STP
D. 1 molar solution of any solute
Correct Answer: B
Explanation: One mole contains $N_A \approx 6.022\times 10^{23}$ entities (atoms, molecules, ions, etc.).
Question 2
How many moles are present in 11 g of $CO_2$? (Molar mass of $CO_2 = 44,g,mol^{-1}$)
A. 0.11
B. 0.25
C. 0.5
D. 2.0
Correct Answer: B
Explanation: $n = \frac{m}{M} = \frac{11}{44} = 0.25$ mol.
Question 3
The molar mass of $CaCO_3$ is:
A. 60 g/mol
B. 80 g/mol
C. 100 g/mol
D. 120 g/mol
Correct Answer: C
Explanation: $CaCO_3 = 40 + 12 + 3\times 16 = 100,g,mol^{-1}$.
Question 4
Which statement is correct about the empirical formula?
A. It shows the actual number of atoms in a molecule.
B. It shows the simplest whole number ratio of atoms.
C. It is always the same as molecular formula.
D. It is used only for ionic compounds.
Correct Answer: B
Explanation: Empirical formula gives the simplest ratio of atoms (e.g., $CH_2O$ for glucose).
Revision Tip: In mole problems, write the conversion chain first: $m\rightarrow n\rightarrow N$ (mass → moles → number of particles).
Question 5
If a compound has empirical formula $CH_2$ and molar mass 42 g/mol, the molecular formula is:
A. $C_2H_4$
B. $C_3H_6$
C. $C_6H_{12}$
D. $C_7H_{14}$
Correct Answer: B
Explanation: Empirical formula mass of $CH_2$ is 14. Factor $= \frac{42}{14} = 3$. So molecular formula $= (CH_2)_3 = C_3H_6$.
Question 6
The percentage by mass of oxygen in water ($H_2O$) is:
A. 11.11%
B. 44.44%
C. 66.67%
D. 88.89%
Correct Answer: D
Explanation: Molar mass of $H_2O = 18$. Oxygen contributes 16. So %O $=\frac{16}{18}\times 100 \approx 88.89%$.
Question 7
In the reaction $2H_2 + O_2 \rightarrow 2H_2O$, if 5 mol $H_2$ reacts with 2 mol $O_2$, the limiting reagent is:
A. $H_2$
B. $O_2$
C. Both are limiting
D. Cannot be determined
Correct Answer: B
Explanation: Ratio needed is $H_2:O_2 = 2:1$. For 2 mol $O_2$, required $H_2 = 4$ mol. Available $H_2 = 5$ mol, so $O_2$ limits.
Question 8
Molarity (M) of a solution is defined as:
A. Moles of solute per kg of solvent
B. Moles of solute per litre of solution
C. Grams of solute per litre of solution
D. Grams of solute per kg of solution
Correct Answer: B
Explanation: $M = \frac{\text{moles of solute}}{\text{volume of solution in L}}$.
Revision Tip: For concentration, be careful with the unit: molarity uses litre of solution; molality uses kg of solvent.
Question 9
How many molecules are present in 0.5 mol of $H_2SO_4$?
A. $3.011\times 10^{23}$
B. $6.022\times 10^{23}$
C. $1.2044\times 10^{24}$
D. $3.011\times 10^{24}$
Correct Answer: A
Explanation: Number of molecules $= nN_A = 0.5\times 6.022\times 10^{23} = 3.011\times 10^{23}$.
Question 10
If 4 g of $NaOH$ is dissolved to make 250 mL solution, the molarity is:
A. 0.1 M
B. 0.2 M
C. 0.4 M
D. 1.0 M
Correct Answer: C
Explanation: Moles of $NaOH = \frac{4}{40} = 0.1$ mol. Volume = 0.25 L. $M = \frac{0.1}{0.25} = 0.4$ M.
Prefer level-wise practice? Try: Some Basic Concepts of Chemistry MCQs (Difficulty-wise).
If you’re preparing for a test, redo the questions you got wrong after reading the explanations.