Changes Around Us: Physical and Chemical – Numerical Problems with Stepwise Solutions
Class 7
Science — Chapter 5: Changes Around Us (Physical & Chemical)
20 Numerical Problems with stepwise solutions — NCERT-aligned for CBSE Class 7.
CBSE Board Examination
Focus: calculations related to mass changes, percent composition, gas production and simple stoichiometric reasoning applied to daily-life changes.
Content Bank — Useful formulas & quick reminders
percent (%) = (part / whole) × 100mass change = final mass − initial mass- For reactions: relate amounts using mole ratios from balanced equations (simple 1:1 examples used here).
- Common atomic masses used where needed: Fe = 55.85, O = 16.00, H = 1.008, C = 12.01, Na = 23.0.
- Practical checks: recoverability after evaporation indicates physical change; formation of gas/precipitate usually indicates chemical change.
Numerical Problems & Stepwise Solutions (Topic-wise)
A. Mass changes in physical and chemical processes (1–6)
Problem 1: A 100 g iron nail is left outside and after rusting its mass becomes 130 g. What is the mass of oxygen combined with iron?
Step 1: Final mass = 130 g, initial mass = 100 g.
Step 2: Mass of oxygen combined = 130 − 100 = 30 g.
Answer: 30 g of oxygen combined with the iron.
Problem 2: A copper vessel of mass 500 g loses 2% of its mass due to wear and tear over years. Calculate mass lost and remaining mass.
Step 1: Mass lost = 2% of 500 g = 0.02 × 500 = 10 g.
Step 2: Remaining mass = 500 − 10 = 490 g.
Answer: Mass lost = 10 g; Remaining = 490 g.
Problem 3: A 20 g copper coin gets tarnished forming a thin layer of copper oxide. If mass increases by 1.5%, find the mass of oxide layer added.
Step 1: Mass increase = 1.5% of 20 g = 0.015 × 20 = 0.30 g.
Answer: Mass increase due to oxide = 0.30 g.
Problem 4: 250 g of a mixture contains 60% water (by mass). If all water evaporates, what mass remains?
Step 1: Mass of water = 60% of 250 = 0.60 × 250 = 150 g.
Step 2: Remaining mass = 250 − 150 = 100 g.
Answer: 100 g remains after evaporation.
Problem 5: A sample of 80 g of fruit juice loses 12% mass on drying. Calculate mass lost and final mass.
Step 1: Mass lost = 12% of 80 = 0.12 × 80 = 9.6 g.
Step 2: Final mass = 80 − 9.6 = 70.4 g.
Answer: Mass lost = 9.6 g; Final mass = 70.4 g.
Problem 6: A sealed bottle contains 200 g of a solution. After heating some liquid evaporates and mass reduces to 170 g. What percentage of mass was lost?
Step 1: Mass lost = 200 − 170 = 30 g.
Step 2: Percentage lost = (30 / 200) × 100 = 15%.
Answer: 15% of mass was lost.
B. Gas evolution & simple stoichiometric reasoning (7–11)
Problem 7: Zinc reacts with dilute HCl to produce hydrogen gas: Zn + 2HCl → ZnCl₂ + H₂. If 65.38 g (1 mol) Zn reacts, find mass of H₂ produced. (H₂ molar mass = 2.016 g/mol)
Step 1: Moles Zn = 65.38 / 65.38 = 1 mol.
Step 2: Stoichiometry: 1 mol Zn → 1 mol H₂.
Step 3: Mass H₂ = 1 × 2.016 = 2.016 g.
Answer: 2.016 g of hydrogen gas.
Problem 8: In vinegar (acetic acid) + baking soda reaction, CO₂ is produced. If reaction gives 44 g of CO₂ (1 mol), how many moles of CO₂ were formed and what mass of sodium bicarbonate (NaHCO₃, molar mass ≈ 84.01 g/mol) was consumed assuming 1:1 mol ratio?
Step 1: Moles CO₂ = 44 / 44.01 ≈ 1.00 mol (approx).
Step 2: If 1 mol CO₂ produced, 1 mol NaHCO₃ consumed (for simple classroom equation).
Step 3: Mass NaHCO₃ ≈ 1 × 84.01 ≈ 84.01 g.
Answer: ≈ 1.00 mol CO₂; NaHCO₃ consumed ≈ 84.01 g.
Problem 9: If 2 g of hydrogen gas is collected from a metal‑acid reaction, how many moles of H₂ is this? (Use H₂ = 2.016 g/mol)
Step 1: Moles H₂ = mass / molar mass = 2.000 / 2.016 ≈ 0.9921 mol.
Answer: ≈ 0.992 mol of H₂.
Problem 10: Carbon dioxide (CO₂) collected turns 100 mL of limewater milky. If 1 mol CO₂ makes 1000 mL limewater milky in this test, estimate moles of CO₂ produced in this experiment (assume linearity for simple estimate).
Step 1: If 1000 mL requires 1 mol, then 100 mL corresponds to 0.1 mol (proportional estimate).
Answer: Approx. 0.10 mol CO₂ (simple classroom estimate).
Problem 11: A gas sample of hydrogen has mass 4.032 g. How many moles of H₂ are present and how many moles of Zn would have reacted (Zn + 2HCl → ZnCl₂ + H₂)?
Step 1: Moles H₂ = 4.032 / 2.016 = 2.000 mol.
Step 2: From reaction, 1 mol Zn → 1 mol H₂ ⇒ moles Zn reacted = 2.000 mol.
Answer: 2.000 mol H₂; Zn reacted = 2.000 mol.
C. Percent composition & concentration related problems (12–15)
Problem 12: A mixture contains 40 g salt and 60 g sand. What is the percent (by mass) of salt in the mixture?
Step 1: Total mass = 40 + 60 = 100 g.
Step 2: Percent salt = (40 / 100) × 100 = 40%.
Answer: Salt is 40% by mass.
Problem 13: A beverage contains 5 g sugar in 100 mL. What is concentration in g per 100 mL and percent w/v?
Step 1: Concentration = 5 g per 100 mL (given).
Step 2: Percent w/v = (5 / 100) × 100 = 5% w/v.
Answer: 5 g/100 mL and 5% w/v.
Problem 14: A steel sample contains 2% carbon by mass. In 5 kg of steel, how much carbon (in grams) is present?
Step 1: Convert 5 kg to grams: 5 kg = 5000 g.
Step 2: Carbon mass = 2% of 5000 = 0.02 × 5000 = 100 g.
Answer: 100 g of carbon.
Problem 15: A 250 g solution contains 5% salt by mass. Calculate mass of salt and solvent (water) in the solution.
Step 1: Mass of salt = 5% of 250 = 0.05 × 250 = 12.5 g.
Step 2: Mass of solvent = 250 − 12.5 = 237.5 g.
Answer: Salt = 12.5 g; Solvent = 237.5 g.
D. Practical classroom estimations & proportional reasoning (16–20)
Problem 16: If 10 g of a metal reacts with acid and produces 0.3 g of hydrogen, what percent of the original metal mass was converted into hydrogen (simple percent by mass)?
Step 1: Percent = (0.3 / 10) × 100 = 3%.
Answer: 3% of the original mass became hydrogen gas (classroom percent calculation).
Problem 17: In a class demo, 250 mL of limewater turns milky when bubbled with CO₂. If 500 mL would have turned the same intensity of milkiness, estimate the moles of CO₂ produced if 1000 mL corresponds to 1 mol in this setup.
Step 1: If 1000 mL = 1 mol, then 500 mL = 0.5 mol and 250 mL = 0.25 mol by proportion.
Answer: For 250 mL: 0.25 mol; for 500 mL: 0.5 mol (proportional classroom estimate).
Problem 18: A sample of 100 g mixture contains 30% component A and 70% component B. If 20 g of component B is removed, what is the new percentage of A in the remaining mixture?
Step 1: Initial mass A = 30% of 100 = 30 g; B = 70 g.
Step 2: After removing 20 g B: new mass B = 70 − 20 = 50 g; total mass = 30 + 50 = 80 g.
Step 3: New % A = (30 / 80) × 100 = 37.5%.
Answer: New percentage of A = 37.5%.
Problem 19: A 15 g sample of substance loses 1.5 g on heating. What fraction and percent of the original sample was lost?
Step 1: Fraction lost = 1.5 / 15 = 0.1.
Step 2: Percent lost = 0.1 × 100 = 10%.
Answer: Fraction = 0.10; Percent = 10%.
Problem 20: A student measures mass before and after an experiment: initial total mass 120 g, after reaction products mass 126 g. Explain what this mass increase indicates in case of a metal reacting with oxygen.
Step 1: Mass increased by 6 g (126 − 120).
Step 2: If a metal reacts with oxygen (e.g., rusting or oxide formation), the metal has combined with oxygen from air, increasing mass by oxygen’s mass.
Answer: Mass increase (6 g) indicates oxygen has combined with the metal to form oxides — typical for chemical changes involving oxidation.
Notes: These numerical examples are tailored for Class 7 NCERT-level understanding — they focus on percent, mass changes, simple mole ideas in classroom demonstrations, and proportional reasoning. Atomic masses used for simple mole conversions: H₂ ≈ 2.016 g/mol, CO₂ ≈ 44.01 g/mol, NaHCO₃ ≈ 84.01 g/mol. Use these problems for practice and small classroom quizzes.