Atoms Molecules Exemplar — Class 9 Science NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 9 Science chapter "Atoms Molecules Exemplar" — 5 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 9 Science chapter "Atoms Molecules Exemplar" — 5 important questions with detailed answers for CBSE board…
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Key Questions Covered:
- Calculate the molar mass of CaCO3 (Ca = 40, C = 12, O = 16).
- How many atoms of oxygen are present in 3.2 g of SO2? (S = 32, O = 16, NA = 6…
- What is the law of constant proportions? Give an example.
- How many molecules are present in 88 g of CO2? (C = 12, O = 16, NA = 6.022 × …
- Define relative atomic mass and molecular mass.
Solutions Summary:
| Question | Status |
|---|---|
| Calculate the molar mass of CaCO3 (Ca = 40, C = 12, O = 16). | ✓ Solved |
| How many atoms of oxygen are present in 3.2 g of SO2? (S … | ✓ Solved |
| What is the law of constant proportions? Give an example. | ✓ Solved |
| How many molecules are present in 88 g of CO2? (C = 12, O… | ✓ Solved |
| Define relative atomic mass and molecular mass. | ✓ Solved |
Showing 5 of 5 questions
Q1: Calculate the molar mass of CaCO3 (Ca = 40, C = 12, O = 16).
Molar mass of CaCO3 = 1(40) + 1(12) + 3(16) = 40 + 12 + 48 = 100 g/mol.
Q2: How many atoms of oxygen are present in 3.2 g of SO2? (S = 32, O = 16, NA = 6.022 × 10^23)
Molar mass of SO2 = 32 + 2(16) = 64 g/mol. Moles of SO2 = 3.2/64 = 0.05 mol. One molecule of SO2 has 2 oxygen atoms. Number of O atoms = 0.05 × 2 × 6.022 × 10^23 = 6.022 × 10^22.
Q3: What is the law of constant proportions? Give an example.
The law states that a chemical compound always contains the same elements in the same proportion by mass. Example: water always contains hydrogen and oxygen in mass ratio 1:8 (2g H to 16g O). No matter the source of water, this ratio remains constant.
Q4: How many molecules are present in 88 g of CO2? (C = 12, O = 16, NA = 6.022 × 10^23)
Molar mass of CO2 = 12 + 2(16) = 44 g/mol. Moles = 88/44 = 2 mol. Number of molecules = 2 × 6.022 × 10^23 = 1.204 × 10^24.
Q5: Define relative atomic mass and molecular mass.
Relative atomic mass: mass of an atom compared to 1/12 of mass of carbon-12 atom. Example: Na = 23 (sodium atom is 23 times heavier than 1/12 of C-12). Relative molecular mass: sum of relative atomic masses of all atoms in a molecule. Example: H2O = 1 + 1 + 16 = 18.
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