### Theory:

 1 Calculate the mass of $$1.51\times10^{23}$$ molecules of water.

Data:

Given molecules of $$H_2O=1.51\times10^{23}$$

Molecular mass of $$H_2O=18$$

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

Solution:

$\begin{array}{lll}\mathit{Number}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{moles}& =& \frac{\mathit{Number}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{molecules}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{water}}{\mathit{Avogadro}\phantom{\rule{0.147em}{0ex}}\mathit{number}}\\ & =& \frac{1.51×{10}^{23}}{6.023×{10}^{23}}\\ & =& \frac{1}{4}\\ & =& 0.25\phantom{\rule{0.147em}{0ex}}\mathit{mole}\end{array}$

$\begin{array}{lll}\mathit{Number}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{moles}& =& \frac{\mathit{Mass}}{\mathit{Molecular}\phantom{\rule{0.147em}{0ex}}\mathit{mass}}\\ 0.25& =& \frac{\mathit{Mass}}{18}\\ \mathit{Mass}& =& 0.25×18\\ \mathit{Mass}& =& 4.5\phantom{\rule{0.147em}{0ex}}g\end{array}$

Hence, the mass of $$1.51\times10^{23}$$ molecules of water $$=4.5$$ $$g$$

 2 Calculate the mass of $$5\times10^{23}$$ molecules of glucose.

Data:

Given molecules of glucose $$=1.51\times10^{23}$$

Molecular mass of glucose $$=180$$

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

Formula:

$\begin{array}{lll}\mathit{Mass}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{glucose}& =& \frac{\mathit{Molecular}\phantom{\rule{0.147em}{0ex}}\mathit{mass}×\mathit{Number}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{particles}}{\mathit{Avogadro}\phantom{\rule{0.147em}{0ex}}\mathit{number}}\\ & =& \frac{180×5×{10}^{23}}{6.023{×10}^{23}}\\ & =& 149.43\phantom{\rule{0.147em}{0ex}}g\end{array}$

Hence, the mass of $$5\times10^{23}$$ molecules of glucose $$=149.43$$ $$g$$