# What is the Relationship anywhere between a Mole and you may Avogadro’s number

Avogadro’s number Atoms and molecules are so small in size that they cannot be counted individually. The chemists use the unit mole for counting atoms, molecules or ions. It is represented by n. A mole represents six.022 ? 10 23 particles. Example: 1 mole of atoms = 6.022 ? 10 23 atoms. 1 mole of molecules = 6.022 ? 10 23 molecules The number of particles present in 1 mole of any substance is fixed i.e. 6.022 ? 10 23 . This number is called Avogadro constant or Oxford sugar daddy dating site Avogadro number. it is represented by No. 1 mole of atoms = 6.022 ? 10 23 atoms = Gram atomic mass or Molar mass of element Number of moles = \(\frac < Mass\quad> < Molar\quad>\) n = \(\frac < m> < M>\) Number of moles = \(\frac < Given\quad> < Avagadro\quad>\) n = \(\frac < N>< <>_ < 0>> \) No. of moles = n Given mass = m Molar mass = M Given number of particles = N Avogadro number of particles = N_{0} These relations can be interchanged as Mass of element, m = n ? M or No. of particles of element, N = n ? N_{0} Similarly, 1 Mole of molecules = 6.022 ? 10 23 molecules = Gram molecular mass of Molar mass Number of moles = \(\frac < Mass\quad> < Molar\quad>\) n = \(\frac < m> < M>\) Number of moles = \(\frac < Given\quad> < Avagadro\quad>\) n = \(\frac < N>< <>_ < 0>> \) or m = n ? M and N = n ? N_{0}

Molarity (M) : Moles of solute is one litre of solution is known as molarity . M = \(\frac < Number\quad> < Volume\quad>\)

Example 1: 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, n = \(\frac < m> < M>\) Mass of silver, m = 20 g, Molar mass of silver, M = 108 g ? n = \(\frac < 20> < 108>\) = 0.185 mol.

## What is the Dating anywhere between a Mole and Avogadro’s number

Example 2: How many moles of CO_{2} are present in 51.2 g of it ? Solution: Molecular mass of CO_{2} = a dozen + 2 + 16 = 44 u Molar mass of CO_{2} (M) = 44 g Mass of CO_{2} (m) = 51.2 g Moles of CO_{2}, n = \(\frac < m> < M>=\frac < 51.2> < 44>\) = 1.16 mol.

Example 3: Calculate the mass of (i) 0.5 moles of N_{2} gas (ii) 0.5 moles of N atoms Solution: (i) 0.5 moles of N_{2} 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

Mass percentage of an element from molecular formula : The molecular formula of a compound may be defined as the formula which specifies the number of atoms of various elements in the molecule of the compound. Example: The molecular formula of glucose is C_{6}H_{12}O_{6}. This indicates that a molecule of glucose contains six atoms of carbon, twelve atoms of hydrogen and six atoms of oxygen. The mass percentage of each element is then calculated by the following formula : Mass percentage of element X = \(\frac < Mass\quad> < Gram\quad>\times 100\)

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