Class 11 Chemistry · Original EZT practice
Some Basic Concepts of Chemistry (Class 11): original worked practice
A balanced equation counts amounts, not grams directly. Keep the substance, the entity being counted and the unit beside each number before converting from mass to moles.
8 original worked examples, 8 fresh self-check questions and 4 mixed questions, with explained answers. These are not copied textbook exercises, an official chapter index or a full answer key. No affiliation with NCERT is claimed.
Official sources for topic and edition context, checked 17 October 2026:
Choose, solve, check, then try a fresh question.
Before calculating, read the method prompt and name the condition that makes it useful. Write your own attempt. Open the steps only when you need them, and use the check to catch a wrong sign, unit or interpretation. Finish with the fresh question before reading its answer.
Include the gas in the mass balance
Choose the method: In a closed system mass is conserved; an open vessel can lose matter to its surroundings.
EZT practice question
A fictional closed reaction system has initial mass 37.5 g. At the end, 31.2 g is non-gaseous material. What mass remains as gas inside the system?
Show the worked solution
- The gas is still part of the closed system.
- Gas mass = 37.5 − 31.2 = 6.3 g.
Answer: 6.3 g
Check: 31.2 + 6.3 = 37.5 g, so the total is unchanged.
Watch for this: A smaller mass of solid or liquid does not mean mass was destroyed.
Now try it yourself
An open vessel loses only gas and its contents fall from 24.8 g to 20.1 g. Find the gas mass lost.
Check your answer
4.7 g
24.8 − 20.1 = 4.7 g under the stated assumption that no other matter enters or leaves.
Name the entity before counting
Choose the method: n = m/M gives moles of the named substance. Multiply by the number of atoms per molecule to count atoms.
EZT practice question
For 13.2 g of CO₂, find the amount of CO₂ and estimate the number of oxygen atoms.
Show the worked solution
- M(CO₂) = 12 + 2 × 16 = 44 g mol⁻¹, so n = 13.2/44 = 0.300 mol.
- Each CO₂ molecule contains two oxygen atoms. Oxygen atoms = 2 × 0.300 × 6.022 × 10²³.
Answer: 0.300 mol CO₂; approximately 3.613 × 10²³ oxygen atoms.
Check: There are 0.600 mol of oxygen atoms but only 0.300 mol of CO₂ molecules.
Watch for this: Do not confuse oxygen atoms bound in CO₂ with free O₂ molecules.
Now try it yourself
Estimate the hydrogen-atom count in 5.4 g of water using the same constants.
Check your answer
Approximately 3.613 × 10²³ hydrogen atoms.
5.4/18 = 0.300 mol H₂O; each molecule has two hydrogen atoms, giving 0.600 N_A atoms.
Apply each subscript to the correct atom
Choose the method: Molar mass adds the relative atomic-mass contributions and has unit g mol⁻¹ for these calculations.
EZT practice question
Find M(Na₂CO₃), then the mass of 0.150 mol of this compound.
Show the worked solution
- M = 2 × 23 + 12 + 3 × 16 = 106 g mol⁻¹.
- m = nM = 0.150 × 106 = 15.9 g.
Answer: 106 g mol⁻¹; 15.9 g
Check: 15.9 g divided by 106 g mol⁻¹ returns 0.150 mol.
Watch for this: The subscript 2 applies to Na, not to every element that follows.
Now try it yourself
Find M(NaHCO₃) and the mass of 0.250 mol.
Check your answer
84 g mol⁻¹; 21.0 g
23 + 1 + 12 + 48 = 84; 0.250 × 84 = 21.0.
Use the whole compound as the denominator
Choose the method: An element's mass percentage is its mass contribution divided by the total formula mass, multiplied by 100.
EZT practice question
Find the nitrogen mass percentage in NH₃, then the nitrogen mass in 8.50 g of NH₃.
Show the worked solution
- The formula mass is 14 + 3 × 1 = 17. Nitrogen contributes 14.
- Nitrogen percentage = (14/17) × 100 ≈ 82.4%.
- Nitrogen mass = 8.50 × 14/17 = 7.00 g.
Answer: Approximately 82.4% nitrogen; 7.00 g nitrogen.
Check: The remaining 1.50 g is hydrogen under the stated atomic-mass model.
Watch for this: Dividing 14 by 3 compares two constituents, not nitrogen with the whole compound.
Now try it yourself
Find the hydrogen percentage and hydrogen mass in 3.40 g NH₃.
Check your answer
Approximately 17.6%; 0.600 g hydrogen.
Hydrogen contributes 3/17 of the mass; 3.40 × 3/17 = 0.600 g.
Convert element masses to amounts first
Choose the method: The empirical formula uses the simplest whole-number atom ratio, not the mass ratio.
EZT practice question
A compound contains only carbon and oxygen. A sample contains 3.6 g carbon and 9.6 g oxygen. Find its empirical formula.
Show the worked solution
- Carbon amount = 3.6/12 = 0.30 mol; oxygen-atom amount = 9.6/16 = 0.60 mol.
- Divide both by 0.30 to obtain C:O = 1:2.
Answer: CO₂
Check: The formula implies mass ratio 12:32 = 3.6:9.6.
Watch for this: Using the raw mass ratio gives the wrong atom ratio.
Now try it yourself
A compound contains only nitrogen and oxygen: 2.8 g nitrogen and 4.8 g oxygen. Find its empirical formula.
Check your answer
N₂O₃
Amounts are 0.20 mol N atoms and 0.30 mol O atoms; the ratio 1:1.5 becomes 2:3.
Use molar mass to scale the empirical unit
Choose the method: For a molecular compound, molecular-formula subscripts are a whole-number multiple of the empirical formula.
EZT practice question
A molecular compound has empirical formula CH and molar mass 78 g mol⁻¹. Find its molecular formula.
Show the worked solution
- The empirical-formula mass is 12 + 1 = 13.
- The multiplier is 78/13 = 6, so multiply both subscripts by 6.
Answer: C₆H₆
Check: 6 × 12 + 6 × 1 = 78, matching the stated molar mass.
Watch for this: Do not add 6 to the subscripts or treat CH as necessarily a separate molecule.
Now try it yourself
A compound has empirical formula NO₂ and molar mass 92 g mol⁻¹. Find its molecular formula.
Check your answer
N₂O₄
The empirical-formula mass is 14 + 32 = 46; 92/46 = 2.
Use final solution volume
Choose the method: Molarity is moles of solute per litre of solution, not per litre of solvent added.
EZT practice question
A written data card gives 0.0750 mol of dissolved solute in a final solution volume of 300 mL. Find its molarity.
Show the worked solution
- 300 mL = 0.300 L.
- M = 0.0750/0.300 = 0.250 mol L⁻¹.
Answer: 0.250 mol L⁻¹
Check: 0.250 mol L⁻¹ × 0.300 L = 0.0750 mol.
Watch for this: The volume refers to the solution after dissolving, not just the solvent before it.
Now try it yourself
The same amount of solute is present after dilution to a final 750 mL, with no reaction or loss. Find the new molarity.
Check your answer
0.100 mol L⁻¹
The amount stays 0.0750 mol and the new solution volume is 0.750 L.
Compare available amounts with coefficients
Choose the method: For aA + bB, compare n(A)/a with n(B)/b. The smaller value sets the reaction extent.
EZT practice question
For the paper model N₂ + 3H₂ → 2NH₃, take 0.40 mol N₂ and 0.90 mol H₂. Assuming complete reaction to the limiting amount, find NH₃ formed and N₂ left.
Show the worked solution
- Compare 0.40/1 with 0.90/3 = 0.30. H₂ limits this model.
- NH₃ formed = 2 × 0.30 = 0.60 mol. N₂ used = 0.30 mol.
- N₂ left = 0.40 − 0.30 = 0.10 mol.
Answer: 0.60 mol NH₃ formed; 0.10 mol N₂ left.
Check: Mass balance: initial 11.2 + 1.8 = 13.0 g; final NH₃ 10.2 g plus N₂ 2.8 g = 13.0 g.
Watch for this: The reactant with fewer moles is not automatically limiting; compare the coefficient-scaled amounts.
Now try it yourself
Use 0.20 mol N₂ and 0.90 mol H₂ in the same complete-reaction model. Find NH₃ formed and H₂ left.
Check your answer
0.40 mol NH₃; 0.30 mol H₂ left.
N₂ limits at extent 0.20 mol, consuming 0.60 mol H₂ and forming 0.40 mol NH₃. Real equilibrium yield is outside this model.
Try a mixed set
Choose a method or explanation before opening each answer.
1. Does a mole always mean a mole of molecules?
Show answer and reason
No.
Specify the entity: atoms, molecules, ions, formula units or another stated group.
2. How does molality differ from molarity?
Show answer and reason
Molality uses moles of solute per kilogram of solvent; molarity uses moles per litre of solution.
The different denominator is essential; temperature-related volume changes can affect molarity for a fixed composition.
3. A pure sample is CO₂. Does its carbon mass percentage double when the sample mass doubles?
Show answer and reason
No.
The carbon mass and total mass scale by the same factor, so their ratio is unchanged.
4. Is 6.022 × 10²³ the exact defining numerical value of N_A?
Show answer and reason
No. It is a rounded working value.
The exact SI value is 6.02214076 × 10²³ mol⁻¹. Distinguish the definition from a school calculation approximation.
Know the rule but lose track of the steps?
Bring your book edition, your school topic list and a question you attempted. Ask a tutor to explain the choice of method and test it with a different problem. Agree the lesson topics, schedule and fee before starting.
For cost context, see home tuition fees in Delhi NCR. Agree the topics, schedule and fee with your chosen tutor.
Questions about this practice page
- What is the current SI definition of the mole?
- One mole contains exactly 6.02214076 × 10²³ specified elementary entities. The older carbon-12 mass account is historical context, not the current defining statement.
- Are the atomic masses here exact physical constants?
- No. They are stated rounded values for these written calculations. Use the values given in your own problem.
- Does the smallest mass identify a limiting reagent?
- No. Convert to amounts and compare them with the balanced-equation coefficients.
- Does this cover the whole unit?
- No. It is selected quantitative practice. Measurement, significant figures and other unit topics need separate study.
- Are these NCERT exercises?
- No. The questions, numbers and checks are original selected practice, not an exercise key or laboratory procedure.