Example 3.4 - Chapter 3 Class 9 - Atoms And Molecules
Last updated at April 16, 2024 by Teachoo
Examples from NCERT Book
Examples from NCERT Book
Last updated at April 16, 2024 by Teachoo
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