Measurement of Matter — Maharashtra (SSC) Class 9 Science Solutions (Free)

Free step-by-step Maharashtra (SSC) Class 9 Science solutions for "Measurement of Matter" — important questions with detailed answers, download PDF for board exam preparation.

Measurement of Matter — Maharashtra (SSC) Class 9 Science Solutions (Free)

Free step-by-step Maharashtra (SSC) Class 9 Science solutions for "Measurement of Matter" — important questions with detailed answers, download PDF for board exam preparation.

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Measurement of Matter — Textbook Questions Answered (7)

Q1: Define relative atomic mass and relative molecular mass.

Relative atomic mass (RAM): Mass of one atom of an element compared to 1/12 of mass of one ¹²C atom. It is a ratio, no units. Relative molecular mass (RMM): Sum of RAM of all atoms in a molecule. Example: H₂O = 2(1) + 16 = 18, so RMM of water is 18.

Q2: Calculate the number of moles in 44 g of CO₂ (C = 12, O = 16).

Step 1: Calculate molar mass of CO₂ = 12 + 2(16) = 12 + 32 = 44 g/mol Step 2: Given mass = 44 g Step 3: Number of moles = mass/molar mass = 44/44 = 1 mole Number of moles = 1 mol

Q3: State the law of conservation of mass with an example.

Law of conservation of mass: Mass cannot be created or destroyed in a chemical reaction; total mass of reactants = total mass of products. Example: Burning 12 g of carbon in 32 g oxygen produces 44 g CO₂. (12 + 32 = 44). This applies to all chemical changes.

Q4: Define Avogadro's number and its value.

Avogadro's number (Nₐ) is the number of particles (atoms, molecules, ions, electrons) present in one mole of a substance. Value: Nₐ = 6.02 × 10²³ particles/mole. Example: 1 mole of H₂ contains 6.02 × 10²³ molecules of H₂.

Q5: Calculate the number of atoms in 4 g of helium (He = 4, Nₐ = 6.02 × 10²³).

Step 1: Molar mass of He = 4 g/mol Step 2: Number of moles = mass/molar mass = 4/4 = 1 mole Step 3: Number of atoms = moles × Nₐ = 1 × 6.02 × 10²³ = 6.02 × 10²³ atoms

Q6: State the law of constant proportions (law of definite proportions).

Law of constant proportions: A chemical compound always contains the same elements combined together in the same proportion by mass, regardless of its source or method of preparation. Example: Water (H₂O) always has 2 parts H to 16 parts O by mass, whether from rain, sea, or laboratory synthesis.

Q7: Calculate the molar mass of Ca(OH)₂ (Ca = 40, O = 16, H = 1).

Step 1: Identify atoms: 1 Ca + 2 O + 2 H Step 2: Molar mass = 1(40) + 2(16) + 2(1) = 40 + 32 + 2 = 74 g/mol Molar mass of Ca(OH)₂ = 74 g/mol

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