### Theory:

 1 Calculate the number of molecules in $$11.2$$ $$L$$ of $$CO_2$$ at STP.

Data:

Given volume of $$CO_2=11.2$$ $$L$$

Molar volume of $$CO_2=22.4$$ $$L$$

Avogadro's number $$=6.023\times10^{23}$$

Solution:

$\begin{array}{lll}\mathit{Number}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{moles}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}{\mathit{CO}}_{2}& =& \frac{\mathit{Volume}\phantom{\rule{0.147em}{0ex}}\mathit{at}\phantom{\rule{0.147em}{0ex}}\mathit{STP}}{\mathit{Molar}\phantom{\rule{0.147em}{0ex}}\mathit{volume}}\\ & =& \frac{11.2}{22.4}\\ & =& 0.5\phantom{\rule{0.147em}{0ex}}\mathit{mole}\end{array}$

$\begin{array}{lll}\mathit{Number}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{molecules}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}{\mathit{CO}}_{2}& =& \mathit{Number}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{moles}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}{\mathit{CO}}_{2}×\mathit{Avogadro}\phantom{\rule{0.147em}{0ex}}\mathit{number}\\ & =& 0.5×6.023×{10}^{23}\\ & =& 3.011×{10}^{23}\phantom{\rule{0.147em}{0ex}}\mathit{molecules}\end{array}$

Therefore, the number of molecules in $$11.2$$ $$L$$ of $$CO_2$$ at STP $$=3.011\times10^{23}$$

 2 Calculate the number of molecules in $$54$$ $$g$$ of $$H_2O$$.

Data:

Given mass of $$H_2O=54$$ $$g$$

Gram molecular mass of $$H_2O=18$$ $$g$$

Avogadro's number $$=6.023\times10^{23}$$

Solution:

$\begin{array}{l}\mathit{Number}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{molecules}=\frac{\mathit{Avogadro}\phantom{\rule{0.147em}{0ex}}\mathit{number}×\mathit{Given}\phantom{\rule{0.147em}{0ex}}\mathit{mass}}{\mathit{Gram}\phantom{\rule{0.147em}{0ex}}\mathit{molecular}\phantom{\rule{0.147em}{0ex}}\mathit{mass}}\\ \begin{array}{lll}\mathit{Number}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{molecules}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{water}\phantom{\rule{0.147em}{0ex}}& =& 6.023×{10}^{23}×\frac{54}{18}\\ & =& 18.069×{10}^{23}\end{array}\end{array}$

Therefore, the number of molecules in $$54$$ $$g$$ of $$H_2O=18.069\times10^{23}$$ $$molecules$$