Class 9
Chapter 3 Class 9 - Atoms And Molecules

Calculate the mass of  (i) 0.5 mole of N2 gas (ii) 0.5 mole of N atoms - Examples from NCERT Book

part 2 - Example 3.4 - Examples from NCERT Book - Chapter 3 Class 9 - Atoms And Molecules - Class 9
part 3 - Example 3.4 - Examples from NCERT Book - Chapter 3 Class 9 - Atoms And Molecules - Class 9
part 4 - Example 3.4 - Examples from NCERT Book - Chapter 3 Class 9 - Atoms And Molecules - Class 9
part 5 - Example 3.4 - Examples from NCERT Book - Chapter 3 Class 9 - Atoms And Molecules - Class 9 part 6 - Example 3.4 - Examples from NCERT Book - Chapter 3 Class 9 - Atoms And Molecules - Class 9

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Example 3.4 Calculate the mass of the following: (i) 0.5 mole of N2 gas (mass from mole of molecule) Atomic Mass of Nitrogen = 14 u Molar Mass of N2 gas = 2 × 14 = 28 g Now, Required Mass = Molar Mass × Number of Moles = 28 × 0.5 = 14 g Example 3.4 Calculate the mass of the following: (ii) 0.5 mole of N atoms (mass from mole of atom) Atomic Mass of Nitrogen = 14 u Molar Mass of N atoms = 14 g Now, Required Mass = Molar Mass × Number of Moles = 14 × 0.5 = 7 g Example 3.4 Calculate the mass of the following: (iii) 3.011 × 1023 number of N atoms (mass from number) Atomic Mass of Nitrogen = 14 u Molar Mass of N atoms = 14 g Now, Number of moles = (𝐺𝑖𝑣𝑒𝑛 𝑀𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑃𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠)/(𝐴𝑣𝑜𝑔𝑎𝑑𝑟𝑜 𝑁𝑢𝑚𝑏𝑒𝑟) = (3.011 ×10^23)/(6.022 × 10^23 ) = 𝟏/𝟐 Therefore, Required Mass = Molar Mass × Number of Moles = 14 × 𝟏/𝟐 = 7 g Example 3.4 Calculate the mass of the following: (iv) 6.022 × 1023 number of N2 molecules (mass from number) Atomic Mass of Nitrogen = 14 u Molar Mass of N2 gas = 2 × 14 = 28 g Now, Number of moles = (𝐺𝑖𝑣𝑒𝑛 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑃𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠)/(𝐴𝑣𝑜𝑔𝑎𝑑𝑟𝑜 𝑁𝑢𝑚𝑏𝑒𝑟) = (6.022 ×10^23)/(6.022 × 10^23 ) = 1 Therefore, Required Mass = Molar Mass × Number of Moles = 28 × 1 = 28 g

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Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 14 years and a teacher from the past 18 years. He teaches Science, Economics, Accounting and English at Teachoo