Answer: \(9.28\times 10^{23}\) atoms
Explanation:
According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number \(6.023\times 10^{23}\) of particles.
General representation of an element is given as: \(_{Z}^{A}\textrm{X}\)
\(_{35}^{79.90}\textrm{Br}\)
where,
Z represents Atomic number
A represents Mass number
X represents the symbol of an element
1 mole of bromine \(Br\) contains=\(6.022\times 10^{23}\) atoms
1.54 moles of bromine \(Br\) contains=\(\frac{6.022\times 10^{23}}{1}\times 1.54=9.28\times 10^{23}\) atoms
Thus \(9.28\times 10^{23}\) atoms are present in 1.54 moles of bromine.
