Theory:

1.  Calculate the number of atoms present in \(1\) gram of gold (atomic mass of \(Au=198\)).
 
Data:
 
Given mass of gold \(=1\) \(g\)
 
Atomic mass of \(Au=198\)
 
Avogadro's number \(=6.023\times10^{23}\)
 
Solution:
 
NumberofatomsofAu=MassofAu×AvogadronumberAtomicmassofAu=1198×6.023×1023=3.042×1021g
 
Therefore, the number of atoms present in \(1\) gram of gold \(=3.042\times10^{21}\) \(g\)
 
2. Calculate the number of atoms of oxygen and carbon in \(5\) \(moles\) of \(CO_2\).
 
Calculating number of oxygen atoms in \(5\) \(moles\) of \(CO_2\):
  • \(1\)  \(mole\) of \(CO_2\) contains \(2\) \(moles\) of oxygen.
  • \(5\)  \(moles\) of \(CO_2\) contain \(10\) \(moles\) of oxygen.
Solution:
 
No.ofatomsofO2=No.ofmolesofO2×Avogadronumber=10×6.023×1023=6.023×1024
 
Therefore, the number of atoms of oxygen in \(5\) \(moles\) of \(CO_2=6.023\times10^{24}\)
  
Calculating the number of carbon atoms in \(5\) \(moles\) of \(CO_2\):
  • \(1\) \(mole\) of \(CO_2\) contains \(1\) \(moles\) of carbon.
  • \(5\) \(moles\) of \(CO_2\) contain \(5\) \(moles\) of carbon.
No.ofatomsofC=No.ofmolesofC×Avogadronumber=5×6.023×1023=3.011×1024
 
Therefore, the number of atoms of carbon in \(5\) \(moles\) of \(CO_2=3.011\times10^{24}\)