which of the following has maximum number of atoms (a) 18g of H2O(b)18g of O2 (c) 18g of CO2 (d) 18 g of CH4
Answers
GIVEN :
(a) 18 g of H₂O (b) 18 g of O₂ (c) 18 g of CO₂ (d) 18 g of CH₄
TO FIND :
The compound having the maximum number of atoms in it.
SOLUTION :
To find the number of atoms in a compound, we first have to find the number of moles present in it.
• The formula for the number of moles in a compound is given as :
Moles = Given weight / Molecular weight
• According to Avogadro's law, one mole of an element or a compound contains 6.022 × 10²³ molecules.
(a) 18 g of H₂O
• Molecular weight of H₂O = (1 × 2 + 16) g = (2 + 16) g = 18 g
• Number of moles in 18 g of H₂O = 18 g / 18 g = 1
• Applying Avogadro's law, 1 mole of H₂O contains 6.022 × 10²³ molecules,
=> 18 g of H₂O contains 6.022 × 10²³ molecules.
• Number of atoms in 1 molecule of H₂O = 2H + 1O = 3 atoms
• Therefore, number of atoms in 6.022 × 10²³ molecules of H₂O = 3 × 6.022 × 10²³ = 18.066 × 10²³ atoms
(b) 18 g of O₂
• Molecular weight of O₂ = (2 × 16) g = 32 g
• Number of moles present in 18 g of O₂ = 18 g / 32 g = 9 / 16
• Applying Avogadro's law, one mole of O₂ contains 6.022 × 10²³ molecules,
=> 9 / 16 moles of O₂ contain (9 / 16) × 6.022 × 10²³ molecules = 3.387 × 10²³ molecules
• One molecule of O₂ contains 2 atoms of oxygen.
• Number of atoms in 3.387 × 10²³ molecules of O₂ = 2 × 3.387 atoms = 6.744 × 10²³ atoms
• Therefore, 18 g of O₂ contains 6.744 × 10²³ number of atoms.
(c) 18 g of CO₂
• Molecular weight of CO₂ = (12 + 2 × 16) g = (12 + 32) g = 44 g
• Number of moles in 18 g of CO₂ = 18 g / 44 g = 9 / 22
• Applying Avogadro's law, one mole of CO₂ contains 6.022 × 10²³ molecules,
=> 9 / 22 moles of CO₂ contains (9 / 22) × 6.022 × 10²³ molecules = 2.463 × 10²³ molecules
• From its molecular formula, one molecule of CO₂ contains 1C + 2O atoms = 3 atoms.
• Therefore, 2.463 × 10²³ molecules of CO₂ contains 3 × 2.463 × 10²³ atoms = 7.389 × 10²³ atoms.
• Therefore, 18 g of CO₂ contains 7.389 × 10²³ number of atoms.
(d) 18 g of CH₄
• Molecular weight of CH₄ = (12 + 4 × 1) g = (12 + 4) g = 16 g
• Number of moles present in 18 g of CH₄ = 18 g / 16 g = 9 / 8
• Applying Avogadro's law, one mole of CH₄ contains 6.022 × 10²³ molecules,
=> 9 / 8 moles of CH₄ contains (9 / 8) × 6.022 × 10²³ molecules = 6.774 × 10²³ molecules
• One molecule of CH₄ contains 1C + 4H atoms = 5 atoms
• Therefore, 6.774 × 10²³ molecules of CH₄ contains 5 × 6.774 × 10²³ atoms = 33.87 × 10²³ atoms
• Comparing the numbers of atoms in all the given compounds, we find :
33.87 × 10²³ (d) > 7.389 × 10²³ (c) > 6.744 × 10²³ (b) > 18.066 × 10²³ (a)
• So, 18 g of CH₄ contains the highest number of atoms.
Answer - Option (d), 18 g of CH₄, contains the maximum number of atoms in it, which is equal to 33.87 × 10²³.