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Question: Calculate the number of calcium ions in \(8.2\)g of Calcium Oxide. [Atomic weight of calcium =\(40\)...

Calculate the number of calcium ions in 8.28.2g of Calcium Oxide. [Atomic weight of calcium =4040]
A.3.44×10203.44 \times {10^{20}} ions
B.3.44×10123.44 \times {10^{12}} ions
C.3.44×10233.44 \times {10^{23}} ions
D.8.810×10218.810 \times {10^{21}} ions

Explanation

Solution

One of a substance can be defined as 6.023×10236.023 \times {10^{23}}ion, atoms, molecules, or sub-atomic particles of any element or a compound. We can calculate the moles using the mole concept and then the number of particles present in it.
Formula Used:
moles=massmolar mass{\text{moles}} = \dfrac{{{\text{mass}}}}{{{\text{molar mass}}}}

Complete step by step answer:
One mole of a compound is equal to its gram molecular weight, one mole of an element is equal to its gram atomic weight, and one mole of an ionic compound is equal to its gram ionic weight.
Hence, the gram ionic weight of Calcium Oxide =40+16=5640 + 16 = 56.
Hence 56 grams of Calcium Oxide = 6.023×10236.023 \times {10^{23}} molecules of calcium oxide
Each molecule of calcium oxide has one calcium ion and an oxygen ion. Hence the total number of ions in 56 grams of Calcium Oxide =2×6.023×10232 \times 6.023 \times {10^{23}}, where 6.023×10236.023 \times {10^{23}} ions are of calcium.
So, 56 grams of Calcium Oxide = 6.023×10236.023 \times {10^{23}} ions are of calcium
Therefore, 8.28.2g of Calcium Oxide:
6.023×102356×8.2=0.88×1023\dfrac{{6.023 \times {{10}^{23}}}}{{56}} \times 8.2 = 0.88 \times {10^{23}}= 8.810×10218.810 \times {10^{21}} ions of calcium.

Hence, the correct answer is option D.

Note:
The concept of mole was developed by Italian Physicist Amedeo Avogadro and hence it is also known as the Avogadro’s number or Avogadro’s constant. The concept was developed to measure particles that were not visible to the naked eyes and hence difficult to weigh and measure. The mole is defined as the number of atoms present in 12 grams of Carbon-12 isotope. Hence the molecular weight of any compound or element and the atomic weight of any element contain one mole of that substance.