Is Matter Around Us Pure? – Numerical Problems with Stepwise Solutions
CBSE Class 9 — Science (Chemistry)
Content Bank — Important Formulas & Definitions
Mass percent (w/w %)
Mass % of solute = (mass of solute / mass of solution) × 100
Mass of solution
mass of solution = mass of solute + mass of solvent
Solubility (NCERT convention)
Solubility = grams of solute that dissolve in 100 g of water (solvent) at given temperature
Dilution relation (mass percent)
m1 × %1 = m2 × %2 (for mass-based dilution where m is mass of solution)
Topic: Concentration — Mass Percent (w/w %)
Q1. If 5 g of sugar is dissolved in 45 g of water, what is the mass percent (w/w %) of sugar in the solution?
Step 1: mass of solute (sugar) = 5 g. mass of solvent (water) = 45 g.
Step 2: mass of solution = 5 + 45 = 50 g.
Step 3: Mass % = (mass of solute / mass of solution) × 100 = (5 / 50) × 100 = 10%.
Answer: 10% (w/w)
Q2. A solution contains 12 g of salt (NaCl) in 200 g of solution. Find the mass percent of salt.
Step 1: mass of solute = 12 g; mass of solution = 200 g (given).
Step 2: Mass % = (12 / 200) × 100 = 6%.
Answer: 6% (w/w)
Q3. How many grams of solute are present in 250 g of a 8% (w/w) solution?
Step 1: mass of solution = 250 g, mass % = 8%.
Step 2: mass of solute = (mass % × mass of solution) / 100 = (8 × 250) / 100 = 20 g.
Answer: 20 g of solute.
Topic: Solubility & Saturation
Q4. The solubility of potassium nitrate (KNO₃) at a given temperature is 40 g per 100 g of water. How much KNO₃ will dissolve in 250 g of water at that temperature?
Step 1: Solubility = 40 g KNO₃ per 100 g water.
Step 2: For 250 g water, amount = 40 × (250 / 100) = 40 × 2.5 = 100 g.
Answer: 100 g of KNO₃ will dissolve in 250 g of water (at that temperature).
Q5. If 80 g of solute dissolves in 200 g of water at 40°C, express its solubility as grams per 100 g of water.
Step 1: Given: 80 g solute per 200 g water.
Step 2: For 100 g water, solubility = 80 × (100 / 200) = 80 × 0.5 = 40 g per 100 g water.
Answer: 40 g per 100 g of water.
Q6. A saturated solution at 30°C contains 36 g of compound X in 90 g of water. Is the solution saturated as per solubility expressed per 100 g water? What is the solubility per 100 g water?
Step 1: Given 36 g solute per 90 g water.
Step 2: For 100 g water, solubility = 36 × (100 / 90) = 36 × 1.111... ≈ 40.0 g.
Answer: Solubility ≈ 40 g per 100 g water — so if the known solubility at 30°C is 40 g/100 g water, the solution is saturated.
Topic: Preparation of Solutions & Dilution
Q7. How would you prepare 150 g of a 10% (w/w) sugar solution starting from pure sugar and water? (Calculate masses required)
Step 1: mass of solution required = 150 g, desired mass % = 10%.
Step 2: mass of sugar (solute) = (10/100) × 150 = 15 g.
Step 3: mass of water (solvent) = 150 − 15 = 135 g.
Answer: 15 g sugar and 135 g water.
Q8. You have 300 g of a 12% (w/w) salt solution. How much water must be added to make it a 6% (w/w) solution?
Step 1: Initial solution: mass = 300 g, % = 12%. So mass of salt = 0.12 × 300 = 36 g (salt remains constant on dilution).
Step 2: Let final mass of solution = M. We need (36 / M) × 100 = 6% ⇒ 36 / M = 0.06 ⇒ M = 36 / 0.06 = 600 g.
Step 3: Water to add = final mass − initial mass = 600 − 300 = 300 g.
Answer: Add 300 g of water.
Q9. From a 20% (w/w) solution, how much of it is required to obtain 500 g of a 5% (w/w) solution? (Assume only dilution with water)
Step 1: Let mass of 20% solution required = m. Mass of solute in it = 0.20 × m.
Step 2: After dilution final mass = 500 g with solute mass unchanged. We want 5% ⇒ solute mass = 0.05 × 500 = 25 g.
Step 3: So 0.20 × m = 25 ⇒ m = 25 / 0.20 = 125 g.
Answer: 125 g of the 20% solution is required (then add water to make total 500 g).
Topic: Purity & Percentage Purity
Q10. An ore contains 20% pure metal by mass. How much ore is needed to obtain 50 g of the pure metal (assuming 100% extraction)?
Step 1: Let mass of ore required = m. Pure metal mass = 0.20 × m = 50 g.
Step 2: m = 50 / 0.20 = 250 g.
Answer: 250 g of ore.
Q11. A sample of impure sodium chloride weighs 250 g. After purification, 225 g of pure NaCl is obtained. Calculate the percentage purity of the sample.
Step 1: percentage purity = (mass of pure substance / mass of impure sample) × 100 = (225 / 250) × 100.
Step 2: Compute: 225 / 250 = 0.9 ⇒ ×100 = 90%.
Answer: 90% pure.
Q12. A 10 g sample of impure copper contains 8 g of pure copper. What is the percentage purity?
Step 1: percentage purity = (8 / 10) × 100 = 80%.
Answer: 80%.
Topic: Separation Techniques — Recovery Calculations
Q13. A mixture of sand and salt is treated with water and filtered to separate sand. If 120 g of the original mixture contained 30 g of salt, what mass of sand is recovered after filtration (assume no loss of sand)?
Step 1: Total mass = 120 g, salt mass = 30 g → sand mass = 120 − 30 = 90 g.
Step 2: After filtration sand is recovered wholly (no loss given) ⇒ 90 g sand.
Answer: 90 g of sand.
Q14. 50 g of an impure sample was dissolved in water and filtered to remove insoluble impurities. After evaporation, 35 g of pure substance was obtained. What percentage of the sample was soluble?
Step 1: Soluble portion recovered = 35 g out of 50 g total.
Step 2: Percentage soluble = (35 / 50) × 100 = 70%.
Answer: 70% of the sample was soluble.
Topic: Mixtures & Composition
Q15. A 150 g mixture contains alcohol and water in ratio 2 : 3 by mass. Find the masses of alcohol and water.
Step 1: Ratio parts = 2 + 3 = 5 parts correspond to 150 g.
Step 2: 1 part = 150 / 5 = 30 g. Alcohol (2 parts) = 2 × 30 = 60 g. Water (3 parts) = 3 × 30 = 90 g.
Answer: Alcohol = 60 g; Water = 90 g.
Q16. A mixture contains 70% of A and 30% of B by mass. If the total mass is 500 g, calculate masses of A and B.
Step 1: mass of A = 0.70 × 500 = 350 g. mass of B = 0.30 × 500 = 150 g.
Answer: A = 350 g; B = 150 g.
Topic: Crystallisation & Yield
Q17. From 500 g of a saturated solution containing 20% (w/w) solute, how much solute can be recovered by complete evaporation of the solvent?
Step 1: Mass of solution = 500 g, solute % = 20% → mass of solute = 0.20 × 500 = 100 g.
Step 2: On complete evaporation (assuming solute does not decompose) recovered solute = 100 g.
Answer: 100 g of solute.
Q18. A solution contains 15 g of solute in 85 g of water. If this solution is cooled and half of the solute crystallises out, what mass of solute remains in solution?
Step 1: Initial solute mass = 15 g. Half crystallises ⇒ crystallised mass = 15 / 2 = 7.5 g.
Step 2: Remaining solute in solution = 15 − 7.5 = 7.5 g.
Answer: 7.5 g remains in solution.
Topic: Classification & Small Calculations
Q19. A 2% (w/w) colloidal suspension contains 2 g of dispersed phase in 98 g of dispersion medium. If total volume is not required, what is the mass of dispersed phase in 490 g of the same suspension?
Step 1: 2% by mass ⇒ 2 g dispersed phase per 100 g suspension.
Step 2: For 490 g suspension, dispersed mass = 2 × (490 / 100) = 2 × 4.9 = 9.8 g.
Answer: 9.8 g of dispersed phase.
Q20. Sea water contains 3.5% salts by mass. How many kilograms of salts are present in 2 tonnes of sea water?
Step 1: 2 tonnes = 2000 kg. Mass of salts = 3.5% of 2000 kg = 0.035 × 2000 = 70 kg.
Answer: 70 kg of salts.