Chapter : 3. Atoms & Molecules
Relationship between mole, number of particles and mass and interconversion of one into the other
Relationship between mole, number of particles and mass and interconversion of one into the other.
Molarity (M) : - Moles of solute is one litre of solution is known as molarity.
Example. An ornament of silver contains 20 g of silver. Calculate the moles of silver present (atomic mass of silver = 180 u)
Solution. Moles of silver,
Mass of silver, m = 20 g,
Molar mass of silver,
M = 108 g
∴ n =
= 0.185 mol.
Example. How many moles of CO2 are present in 51.2 g of it ?
Solution. Molecular mass of CO2 = 12 + 2 + 16 = 44 u
Molar mass of CO2 (M) = 44 g
Mass of CO2 (m) = 51.2 g
Moles of CO2,
Example. Calculate the mass of
(i) 0.5 moles of N2 gas
(ii) 0.5 moles of N atoms
Solution. (i) 0.5 moles of N2 gas
Mass = Molar mass × Number of moles
m = M × n
M = 28 g, n = 0.5
m = 28 × 0.5 = 14 g
(ii) Mass = Molar mass × Number of moles
m = M × n
n = 0.5 mole, M = 14 g
m = 14 × 0.5 = 7 g
Molarity (M) : - Moles of solute is one litre of solution is known as molarity.
Example. An ornament of silver contains 20 g of silver. Calculate the moles of silver present (atomic mass of silver = 180 u)
Solution. Moles of silver,
Mass of silver, m = 20 g,
Molar mass of silver,
M = 108 g
∴ n =
= 0.185 mol. Example. How many moles of CO2 are present in 51.2 g of it ?
Solution. Molecular mass of CO2 = 12 + 2 + 16 = 44 u
Molar mass of CO2 (M) = 44 g
Mass of CO2 (m) = 51.2 g
Moles of CO2,
Example. Calculate the mass of
(i) 0.5 moles of N2 gas
(ii) 0.5 moles of N atoms
Solution. (i) 0.5 moles of N2 gas
Mass = Molar mass × Number of moles
m = M × n
M = 28 g, n = 0.5
m = 28 × 0.5 = 14 g
(ii) Mass = Molar mass × Number of moles
m = M × n
n = 0.5 mole, M = 14 g
m = 14 × 0.5 = 7 g
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