Properties of Matter & Measurement — IISER Chemistry MCQs with Solutions
Free IISER 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
Assertion (A): The mass of an object is the same on the Earth and on the Moon, but its weight is less on the Moon.
Reason (R): Mass is the amount of matter in a body and is constant, whereas weight is the gravitational force on the body and depends on the local value of gravity.
Choose the correct option.
A. Both A and R are true and R is the correct explanation of A ✓ Correct
B. A is true but R is false
C. Both A and R are true but R is NOT the correct explanation of A
D. A is false but R is true
Solution: Mass is constant everywhere, so it is unchanged from Earth to Moon. Weight is the gravitational force on the body, and the Moon’s weaker gravity makes the weight smaller there. R states exactly why this happens, so both are true and R correctly explains A.
Q2 — Properties of Matter & Measurement · medium · theory
Which of the following correctly pairs a base physical quantity with its SI base unit?
A. Amount of substance — mole ✓ Correct
B. Thermodynamic temperature — degree Celsius
C. Luminous intensity — lux
D. Electric current — coulomb
Solution: The mole (mol) is the SI base unit of amount of substance. The SI base unit of thermodynamic temperature is the kelvin (not degree Celsius), of luminous intensity the candela (not lux), and of electric current the ampere (not coulomb).
Q3 — Properties of Matter & Measurement · medium · numerical
Object X has a mass of 40 g and a volume of 5 cm³; object Y has a mass of 90 g and a volume of 15 cm³. The ratio of the density of X to that of Y is:
A. 4 : 3 ✓ Correct
B. 3 : 4
C. 2 : 3
D. 9 : 16
Solution: Density of X = 40 ÷ 5 = 8 g cm⁻³; density of Y = 90 ÷ 15 = 6 g cm⁻³. The ratio = 8 : 6 = 4 : 3. Comparing only masses (40 : 90) or only volumes gives the wrong ratios.
Q4 — Properties of Matter & Measurement · medium · numerical
The lowest possible temperature (absolute zero) is 0 K. Using K = °C + 273.15, this corresponds on the Celsius scale to:
A. 0 °C
B. 273.15 °C
C. −273.15 °C ✓ Correct
D. −100 °C
Solution: Rearranging K = °C + 273.15 gives °C = K − 273.15 = 0 − 273.15 = −273.15 °C. This is why negative temperatures are possible on the Celsius scale but impossible on the kelvin scale.
Q5 — Properties of Matter & Measurement · medium · numerical
Equal volumes of two liquids of densities 1.2 g cm⁻³ and 0.8 g cm⁻³ are mixed. If the volumes are additive and there is no reaction, the density of the mixture is:
A. 0.96 g cm⁻³
B. 2.0 g cm⁻³
C. 1.0 g cm⁻³ ✓ Correct
D. 1.2 g cm⁻³
Solution: Take each volume as V. Total mass = 1.2V + 0.8V = 2.0V; total volume = 2V. Density = 2.0V ÷ 2V = 1.0 g cm⁻³ — the simple average, which is valid here only because the two volumes are equal.
Q6 — Properties of Matter & Measurement · hard · numerical
Equal masses of two liquids of densities 2 g cm⁻³ and 3 g cm⁻³ are mixed. If the volumes are additive, the density of the mixture is:
A. 5 g cm⁻³
B. 2.5 g cm⁻³
C. 2.4 g cm⁻³ ✓ Correct
D. 1.2 g cm⁻³
Solution: Take each mass as m. Volumes are m/2 and m/3, so total volume = m/2 + m/3 = 5m/6 and total mass = 2m. Density = 2m ÷ (5m/6) = 12/5 = 2.4 g cm⁻³. It is NOT the simple average 2.5 g cm⁻³, because the two volumes are unequal.
Q7 — Properties of Matter & Measurement · easy · numerical
How many small cubes of edge 1 cm (each of volume 1 cm³) are needed to exactly fill a vessel of volume 1 dm³?
A. 100
B. 1000000
C. 1000 ✓ Correct
D. 10
Solution: Since 1 dm = 10 cm, a volume of 1 dm³ = (10 cm)³ = 1000 cm³ (which also equals 1 L). As each small cube is 1 cm³, exactly 1000 of them fill 1 dm³.