a) Calculate the molar mass of H${}_{2}$O:
${\mathrm{mm}}_{{H}_{2}\mathrm{O}}equals;2\cdot {\mathrm{mm}}_{H}plus;1\cdot {\mathrm{mm}}_{\mathrm{O}}\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}{\mathrm{mm}}_{{H}_{2}\mathrm{O}}equals;2\cdot 1.01\frac{g}{\mathrm{mol}}plus;1\cdot 16.00\frac{g}{\mathrm{mol}}\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}{\mathrm{mm}}_{{H}_{2}\mathrm{O}}equals;18.02\frac{g}{\mathrm{mol}}$
b) Calculate the number of H${}_{2}$O molecules in 5.0 g of water:
* First determine the amount of substance *
${n}_{{H}_{2}\mathrm{O}}equals;\frac{{m}_{{H}_{2}\mathrm{O}}}{{\mathrm{mm}}_{{H}_{2}\mathrm{O}}}\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\mathit{}{n}_{{H}_{2}\mathrm{O}}equals;\frac{5.0g}{18.02\frac{g}{\mathrm{mol}}}\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}\mathit{}{n}_{{H}_{2}\mathrm{O}}equals;0.28\mathrm{mol}$
* Now determine the number of molecules as shown in the previous example *
$\#{\mathrm{of\; H}}_{2}\mathrm{O}\mathrm{molecules}equals;\mathrm{amount}\mathrm{of}\mathrm{substance}\cdot \mathrm{Avogadro}apos;s\mathrm{constant}\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}num;{\mathrm{of\; H}}_{2}\mathrm{O}\mathrm{molecules}equals;0.277\mathrm{mol}\cdot \left(6.022\cdot {10}^{23}\right){\mathrm{mol}}^{-1}\phantom{\rule[-0.0ex]{0.0em}{0.0ex}}num;{\mathrm{of\; H}}_{2}\mathrm{O}\mathrm{molecules}equals;1.67\cdot {10}^{23}$