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Question: The number of iodine atoms present in \[1{\text{c}}{{\text{m}}^{\text{3}}}\] of its \[0.1\] solution...

The number of iodine atoms present in 1cm31{\text{c}}{{\text{m}}^{\text{3}}} of its 0.10.1 solution is:
A.6.02×10236.02 \times {10^{23}}
B.6.02×10226.02 \times {10^{22}}
C.6.02×10196.02 \times {10^{19}}
D.1.204×10201.204 \times {10^{20}}

Explanation

Solution

In this question, we have to calculate the number of entities in a volume of iodine. Simply use the unitary method to find the answer to this question. 6.022×10236.022 \times {10^{23}} is Avogadro's number and represents the number of atoms/molecules present in one mole of the substance.

Complete step by step answer:
We have to calculate the no. of iodine atoms in 1cm31{\text{c}}{{\text{m}}^{\text{3}}} of its 0.10.1 solution
Firstly, we must know the meaning of 0.10.1 the solution. It means a 0.1M solution that is 0.10.1 moles that are dissolved in 1L1{\text{L}} of solution.
Using the unitary method, we have, 1L→0.1mole1{\text{L}} \to 0.1{\text{mole}}.
Volume is given in the question = 1cm3=10−3L1{\text{c}}{{\text{m}}^{\text{3}}} = {10^{ - 3}}{\text{L}}.
Hence, moles present in 10−3L{10^{ - 3}}{\text{L}}=10−3×0.1moles=10−4moles{10^{ - 3}} \times 0.1{\text{moles}} = {10^{ - 4}}{\text{moles}}.
No. of atoms present in 1 mole→6.022×1023atoms6.022 \times {10^{23}}{\text{atoms}}.
No. of atoms present in 10−4moles{10^{ - 4}}{\text{moles}}= 10−4×6.022×1023=6.022×1019atoms{10^{ - 4}} \times 6.022 \times {10^{23}} = 6.022 \times {10^{19}}{\text{atoms}}.
Hence, the number of iodine atoms present in 1cm31{\text{c}}{{\text{m}}^{\text{3}}} its 0.10.1 solution =6.022×1019atoms6.022 \times {10^{19}}{\text{atoms}}.

Hence, the correct option is option C.

Note:
The concept that a mole of any substance contains the same number of particles was formed out of research which was conducted by Italian physicist Amedeo Avogadro. Avogadro constant can be defined as the number of molecules, atoms, or ions in one mole of a substance: 6.022×10236.022 \times {10^{23}} per mol. It is derived from the number of atoms of the pure isotope 12C^{{\text{12}}}{\text{C}}in 12 grams of that substance and is the reciprocal of atomic mass in grams. The formulae for the mole concept can be summarized as:
No. of moles = Mass of the Substance in gramsMolar mass of a Substance=Number of Atoms or Molecules6.022×1023{\text{No}}{\text{. of moles = }}\dfrac{{{\text{Mass of the Substance in grams}}}}{{{\text{Molar mass of a Substance}}}} = \dfrac{{{\text{Number of Atoms or Molecules}}}}{{6.022 \times {{10}^{23}}}}.
These formulae can be used to establish relationships between formula mass, the mole, and Avogadro’s number can be applied to compute various quantities that describe the composition of substances and compounds.