Answer:
1.593 grams is the mass of \(4.92\times 10^{21}\) platinum atoms.
Explanation:
\(N=n\times N_A\)
\(m=n\times M\)
Where:
N = Number of particles / atoms/ molecules
n = Number of moles
m = mass of sample
M = Atomic mass
\(N_A=6.022\times 10^{23} mol^{-1}\) = Avogadro's number
We have:
Atoms of platinum = N = \(4.92 \times 10^{21}\)
Moles of platinum = n =?
\(n=\frac{N}{N_A}=\frac{4.92\times 10^{21}}{6.022\times 10^{23} mol^{-1}}\)
n = 0.008170 moles
Mass of N platinum atoms = m
Atomic mass of the platinum atom = m = 195 g/mol
\(m=0.008170 mol\times 195 g/mol=1.593 g\)
1.593 grams is the mass of \(4.92\times 10^{21}\) platinum atoms.
