Some Basic Concepts of Chemistry JEE Main PYQs (2020–2026)

Solved Some Basic Concepts of Chemistry PYQs

3 recent questions with answers. The full set of 198 is free to practise.

1. JEE Main 2026EasyPercentage Composition, Empirical and Molecular Formulae
   Q1

   An oxide of iron contains $69.9\%$ iron, its empirical formula, is :
   
   (Given : Molar mass of Fe and O are 56 and $16g{mol}^{-1}$ respectively.)

   • A$FeO$
   • B${Fe}_2{O}_3$
   • C${Fe}_3{O}_4$
   • D${FeO}_3$

   Show answer & solution

   Correct answer: (B)

   Let the oxide of iron contain 100 g of the compound.
   
   Then,
   
   Mass of Fe = 69.9 g
   
   Mass of O = 100 − 69.9 = 30.1 g
   
   Calculate moles of each element
   
   $$\text{Moles of Fe}=\frac{69.9}{56}=1.248$$
   
   $$\text{Moles of O}=\frac{30.1}{16}=1.881$$
   
   Find the simplest whole-number ratio
   
   Divide each by the smaller number (1.248):
   
   $$\text{Ratio of Fe:O}=\frac{1.248}{1.248}:\frac{1.881}{1.248}=1:1.51$$
   
   This ratio is approximately 1 : 1.5.
   
   To remove the fraction, multiply both by 2 : ${\text{Fe}}_2{\text{O}}_3$
   
   In $F{e}_2{O}_3$ ,
   
   Mass of Fe = $2\times 56=112$ g
   
   Mass of O = $3\times 16=48$ g
   
   Total mass = 160 g
   
   Percentage of Fe = $\frac{112}{160}\times 100=70\%$ (approximately 69.9%)
   
   Therefore, the empirical formula of the oxide is
   
   $$\boxed{F{e}_2{O}_3}$$
   
   Correct Option: B.
2. JEE Main 2026EasyConcentration of Solutions (Molarity, Molality, Normality, Mole Fraction)
   Q2

   What is the mole fraction of water in $10\%$ by weight $(w/w)$ of aqueous urea solution?
   
   [Given: Molar mass of $H,O,C$ and N are $1,16,12$ and $14g{mol}^{-1}$ respectively.]

   • A0.825
   • B0.032
   • C0.867
   • D0.967

   Show answer & solution

   Correct answer: (D)

   For a $10\%$ by weight $(w/w)$ aqueous urea solution, let us take $100\text{g}$ of solution.
   
   Then:
   
   Mass of urea $=10\text{g}$
   
   Mass of water $=90\text{g}$
   
   Now find the molar masses.
   
   Urea has formula $N{H}_{2}CON{H}_2$ or $CO(N{H}_2{)}_2$
   
   $$M(\text{urea})=12+16+2(14)+4(1)=60{\text{g mol}}^{-1}$$
   
   Moles of urea:
   
   $$n_{\text{urea}}=\frac{10}{60}=\frac{1}{6}\approx 0.167$$
   
   Molar mass of water:
   
   $$M(\text{water})=2(1)+16=18{\text{g mol}}^{-1}$$
   
   Moles of water:
   
   $$n_{\text{water}}=\frac{90}{18}=5$$
   
   Now, mole fraction of water is
   
   $$X_{\text{water}}=\frac{n_{\text{water}}}{n_{\text{water}}+n_{\text{urea}}}$$
   
   $$X_{\text{water}}=\frac{5}{5+\frac{1}{6}}=\frac{5}{\frac{31}{6}}=\frac{30}{31}$$
   
   $$X_{\text{water}}\approx 0.967$$
   
   So, the correct answer is:
   
   $$\boxed{0.967}$$
   
   Option D
3. JEE Main 2026MediumMole Concept, Molar Mass and Avogadro Number
   Q3

   The correct order of total number of atoms present in
   
   2 moles of cyclohexane
   
   684 g of sucrose
   
   90.8 L of dihydrogen at STP
   
   is :

   • A$C>A>B$
   • B$C>B>A$
   • C$B>C>A$
   • D$B>A>C$

   Show answer & solution

   Correct answer: (D)

   Let us find the total number of atoms in each case.
   
   We only need to compare, so first calculate the number of moles of atoms in each sample.
   
   2 moles of cyclohexane
   
   Cyclohexane has formula:
   
   $$C_6{H}_{12}$$
   
   So, one molecule of cyclohexane contains:
   
   $$6+12=18\text{ atoms}$$
   
   Therefore, in $2$ moles of cyclohexane, moles of atoms are:
   
   $$2\times 18=36\text{ moles of atoms}$$
   
   684 g of sucrose
   
   Sucrose has formula:
   
   $$C_{12}{H}_{22}{O}_{11}$$
   
   Total atoms in one molecule:
   
   $$12+22+11=45$$
   
   Molar mass of sucrose:
   
   $$12\times 12+22\times 1+11\times 16$$
   
   $$=144+22+176=342{\text{ g mol}}^{-1}$$
   
   Given mass $=684$ g, so moles of sucrose:
   
   $$\frac{684}{342}=2\text{ mol}$$
   
   Hence, moles of atoms:
   
   $$2\times 45=90\text{ moles of atoms}$$
   
   90.8 L of dihydrogen at STP
   
   At STP, $1$ mole of any gas occupies $22.4$ L.
   
   So moles of $H_2$:
   
   $$\frac{90.8}{22.4}\approx 4.05\text{ mol}$$
   
   Each molecule of $H_2$ has $2$ atoms.
   
   Therefore, moles of atoms:
   
   $$4.05\times 2=8.10\text{ moles of atoms}$$
   
   Comparison
   
   $=36$ moles of atoms
   
   $=90$ moles of atoms
   
   $=8.10$ moles of atoms
   
   So the correct order is:
   
   $$B>A>C$$
   
   Therefore, the correct option is:
   
   $$\boxed{\text{Option D}}$$

Topic-wise question count

Where JEE Main has asked from Some Basic Concepts of Chemistry, most frequent first.

• Stoichiometry, Stoichiometric Calculations & Limiting Reagent71
• Concentration of Solutions (Molarity, Molality, Normality, Mole Fraction)59
• Percentage Composition, Empirical and Molecular Formulae32
• Mole Concept, Molar Mass and Avogadro Number22
• Laws of Chemical Combinations, Dalton's Theory & Significant Figures7
• Equivalent Concept and Gram Equivalents7

Year-wise Some Basic Concepts of Chemistry questions in JEE Main

Number of questions from this chapter in each year's paper(s).