Answer:
[tex]L \approx 3.103\times 10^{-10}\,m[/tex]
Explanation:
The mass of water is:
[tex]m_{H_{2}O} = (4\,mol)\cdot (18\,\frac{g}{mol} )[/tex]
[tex]m_{H_{2}O} = 72\,g[/tex]
[tex]V_{H_{2}O} = \frac{72\,g} {1\,\frac{g}{cm^{3}} }[/tex]
[tex]V_{H_{2}O}= 72\,cm^{3}[/tex]
According to the Avogadro's Number, there are [tex]6.022\times 10^{23}\,\frac{molecules}{mol}[/tex]. The total number of molecules in the water is:
[tex]n = 2.409\times 10^{24}\,molecules[/tex]
The volume occupied by a molecule is:
[tex]V_{molecule} = \frac{V_{H_{2}O}}{n}[/tex]
[tex]V_{molecule} = \frac{72\times 10^{-6}\,m^{3}}{2.409\times 10^{24}}[/tex]
[tex]V_{molecule} = 2.989\times 10^{-29}\,m^{3}[/tex]
The length of an edge of such cube is:
[tex]L = \sqrt[3]{V_{molecule}}[/tex]
[tex]L \approx 3.103\times 10^{-10}\,m[/tex]
