Properties of Matter & Measurement — JEE Main Chemistry MCQs with Solutions
Free JEE Main Chemistry Properties of Matter & Measurement MCQs with step-by-step solutions (7 questions). Part of Some Basic Concepts of Chemistry. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Properties of Matter & Measurement · medium · theory
Which of the following is NOT one of the seven SI base quantities?
A. Luminous intensity
B. Volume ✓ Correct
C. Electric current
D. Amount of substance
Solution: The seven SI base quantities are length, mass, time, electric current, thermodynamic temperature, amount of substance and luminous intensity. Volume is a derived quantity — it has units of (length)³ — and is therefore not a base quantity.
Q2 — Properties of Matter & Measurement · medium · theory
Which of the following processes involves only a change in physical properties, with NO chemical change?
A. Burning of a magnesium ribbon
B. Melting of ice into water ✓ Correct
C. Rusting of iron
D. Souring of milk into curd
Solution: Melting of ice is a change of physical state; the substance remains H₂O, so its identity and composition are unchanged — a physical change. Rusting, burning and souring all produce new substances with new compositions, so they are chemical changes.
Q3 — Properties of Matter & Measurement · medium · numerical
A metal cube of edge 2 cm has a mass of 72 g. Using density = mass ÷ volume, its density is:
A. 18 g cm⁻³
B. 9 g cm⁻³ ✓ Correct
C. 36 g cm⁻³
D. 4.5 g cm⁻³
Solution: Volume of the cube = (edge)³ = (2 cm)³ = 8 cm³. Density = 72 g ÷ 8 cm³ = 9 g cm⁻³. Using area (2² = 4) instead of volume gives 18; using the edge itself (2) gives 36.
Q4 — Properties of Matter & Measurement · hard · numerical
The density of aluminium is 2.7 g cm⁻³. Expressed in SI units (kg m⁻³), this density is:
A. 2700 kg m⁻³ ✓ Correct
B. 27 kg m⁻³
C. 270000 kg m⁻³
D. 0.0027 kg m⁻³
Solution: Since 1 g = 10⁻³ kg and 1 cm³ = 10⁻⁶ m³, we have 1 g cm⁻³ = 10⁻³ ÷ 10⁻⁶ = 10³ kg m⁻³. Therefore 2.7 g cm⁻³ = 2.7 × 1000 = 2700 kg m⁻³.
Q5 — Properties of Matter & Measurement · medium · numerical
A flask contains 250 g of a solution whose density is 1.25 g mL⁻¹. The volume of the solution is:
A. 500 mL
B. 200 mL ✓ Correct
C. 312.5 mL
D. 125 mL
Solution: Volume = mass ÷ density = 250 g ÷ 1.25 g mL⁻¹ = 200 mL. Multiplying instead (250 × 1.25) gives the wrong 312.5 mL.
Q6 — Properties of Matter & Measurement · hard · numerical
At what Celsius temperature does the Fahrenheit reading become exactly twice the Celsius reading? (Use °F = (9/5)°C + 32.)
A. 160 °C ✓ Correct
B. 80 °C
C. 320 °C
D. 40 °C
Solution: Set °F = 2 × °C: 2C = (9/5)C + 32 ⇒ 2C − 1.8C = 32 ⇒ 0.2C = 32 ⇒ C = 160 °C. Check: °F = 1.8 × 160 + 32 = 320 = 2 × 160. (320 is the Fahrenheit reading, not the Celsius answer.)
Q7 — Properties of Matter & Measurement · medium · numerical
The volume of a container is 0.05 m³. Using 1 m³ = 1000 L, this volume in litres is:
A. 50000 L
B. 500 L
C. 50 L ✓ Correct
D. 5 L
Solution: 1 m³ = 10⁶ cm³ and 1 L = 1000 cm³, so 1 m³ = 1000 L. Therefore 0.05 m³ = 0.05 × 1000 = 50 L. Using a factor of 10⁴ or 10² gives the wrong 500 L or 5 L.