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Question

Question: How many atoms are in \(1.54\)moles of bromine \((Br)\)?...

How many atoms are in 1.541.54moles of bromine (Br)(Br)?

Explanation

Solution

The question is based on the concept of number of moles and the Avogadro Law. According to the Avogadro law 11mole of every substance occupies 22.4L22.4Lof volume at STP and contains 6.023×10236.023\times {{10}^{23}}number of molecules/particles. Hence Avogadro law gives the value of the number of molecules present in 11mole of a substance. So, in order to find the number of molecules in more than one mole we will simply perform the unitary method. Once the number of molecules is calculated then the number of atoms can be calculated by simply multiplying the number of molecules by its atomicity.

Complete step-by-step answer:
By using the principle of Avogadro law, we can say that
Number of molecules in mole of Bromine (Br)=6.023×1023(Br)\,=\,6.023\times {{10}^{23}}
Therefore, number of molecules present in 1.541.54moles of Bromine (Br)(Br)will be given by
1.54×6.023×1023 9.28×1023molecules  \Rightarrow \,1.54\,\times \,6.023\times \,{{10}^{23}} \\\ \Rightarrow \,9.28\,\times \,{{10}^{23}}\,molecules \\\
Now we know that atomicity of the Bromine molecule (Br)(Br)is 22. It means one molecule of bromine contains 22atoms as it is a diatomic molecule.
\because Number of atoms in 11molecule of Bromine = 22atoms
\therefore Number of atoms in 9.28×10239.28\,\times \,{{10}^{23}}molecules = 9.28×1023×29.28\,\times \,{{10}^{23}}\times \,2= 18.56×102318.56\,\times \,{{10}^{23}}atoms

Hence 1.541.54moles of Bromine (Br)(Br)contains 18.56×102318.56\,\times \,{{10}^{23}}atoms.

Note: Do not commit mistakes by taking the number of molecules as number of atoms. This is the most misunderstood fact. By this misconception of atoms and molecules sometimes students leave the question in the mid-way and do not solve the entire question. For calculating the number of atoms in an element always multiply the number of molecules with the atomicity of the element. Perform the mathematical calculations carefully.