### 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:

$\begin{array}{lll}\mathit{Number}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{atoms}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{Au}& =& \frac{\mathit{Mass}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{Au}×\mathit{Avogadro}\phantom{\rule{0.147em}{0ex}}\mathit{number}}{\mathit{Atomic}\phantom{\rule{0.147em}{0ex}}\mathit{mass}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{Au}}\\ & =& \frac{1}{198}×6.023×{10}^{23}\\ & =& 3.042×{10}^{21}\phantom{\rule{0.147em}{0ex}}g\end{array}$

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:

$\begin{array}{lll}\mathit{No}.\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{atoms}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}{O}_{2}& =& \mathit{No}.\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{moles}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}{O}_{2}×\mathit{Avogadro}\phantom{\rule{0.147em}{0ex}}\mathit{number}\phantom{\rule{0.147em}{0ex}}\\ & =& 10×{6.023×10}^{23}\\ & =& {6.023×10}^{24}\end{array}$

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.
$\begin{array}{lll}\mathit{No}.\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{atoms}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}C& =& \mathit{No}.\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{moles}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}C×\mathit{Avogadro}\phantom{\rule{0.147em}{0ex}}\mathit{number}\phantom{\rule{0.147em}{0ex}}\\ & =& 5×{6.023×10}^{23}\\ & =& {3.011×10}^{24}\end{array}$

Therefore, the number of atoms of carbon in $$5$$ $$moles$$ of $$CO_2=3.011\times10^{24}$$