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Convert into mole.
(a) 12 g of oxygen gas
(b) 20 g of water
(c) 22 g of carbon dioxide.

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Last updated date: 29th Mar 2024
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MVSAT 2024
Answer
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Hint: The idea of mole is used to count entities at the microscopic level, i.e., atoms, molecules, particles, ions, electrons, etc. Mole (having symbol mol) was introduced as seventh base quantity for the amount of substance in the SI system.

Complete step by step solution:
One mole is defined as the amount of a substance that contains as many particles or entities as there are atoms in exactly 12g of C-12 isotope. The number of entities in 1 mole is so important that it has been given a separate name, known as Avogadro constant (denoted as Na). The numeric value of this constant is \[6.022\times {{10}^{23}}\].
After defining the mole, it has become easier to know the mass of one mole of a substance. The mass of one mole of a substance in grams is called molar mass of that substance.
Molecular mass is calculated by taking the sum of all the atomic masses of the elements present in the molecule.
Since, we have understood the concept of mole and molecular mass, now we can easily calculate the number of moles present in a given amount (in grams) of a substance.
(a)- 32 g of oxygen gas (\[{{O}_{2}}\] molecule) = 1 mol
       This implies, 1 g of \[{{O}_{2}}\] = \[\dfrac{1}{32}\]mol
       Therefore, 12 g of \[{{O}_{2}}\] = \[\dfrac{1}{32}\times 12\]
                                           = 0.375mol

(b)- 18 g of water (\[{{H}_{2}}O\]) = 1 mol
       1 g of water = \[\dfrac{1}{18}\]mol
       Therefore, 20 g of water = \[\dfrac{1}{18}\times 20\]
                                               =1.111 mol

(c)- 44 g of carbon dioxide (\[C{{O}_{2}}\]) = 1 mol
                  1 g of \[C{{O}_{2}}\] will have = \[\dfrac{1}{44}\] mol
                  Therefore, 22 g of \[C{{O}_{2}}\] = \[\dfrac{1}{44}\times 22\]
                                                         = 0.5 mol.

Note: If we are interested in calculating the number of moles of a particular entity, say A, dissolved in a solution of an entity B (A+B), we can do so by using the concept of mole fraction, which is given as:
\[\text{Mole}\,\text{fraction}\,\text{=}\,\dfrac{\text{moles}\,\text{of}\,\text{substance}}{\text{total}\,\text{number}\,\text{of}\,\text{moles}}\]
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