## Chemistry: Molecular Approach (4th Edition)

(a)$\underline{1.6\times {{10}^{4}}\text{ n}{{\text{m}}^{3}}}$ (b)$\underline{1.3\times {{10}^{-24}}\text{ g}}$ (c)$\underline{1.7\times {{10}^{2}}\text{ g}}$ (d)$\underline{1.3\times {{10}^{20}}}$ (e)$\underline{2.1\text{ L}}$
(a) The volume of nanocontainer is as follows: $V={{a}^{3}}$ Substitute $25\text{ nm}$ for a: \begin{align} & V={{\left( 25\text{ nm} \right)}^{3}} \\ & =15625\text{ n}{{\text{m}}^{3}} \end{align} Round off the answer to $1.6\times {{10}^{4}}\text{ n}{{\text{m}}^{3}}$. The volume of one nanocontainer is $\underline{1.6\times {{10}^{4}}\text{ n}{{\text{m}}^{3}}}$. (b) Convert volume into meters as follows: \begin{align} & V=1.6\times {{10}^{4}}\text{ n}{{\text{m}}^{3}}{{\left( \frac{{{10}^{-9}}\text{ m}}{1\text{ nm}} \right)}^{3}} \\ & =1.6\times {{10}^{-23}}\text{ }{{\text{m}}^{3}} \end{align} Now, convert the density into gram per cubic meter. \begin{align} & d=85\text{ g/L}\left( \frac{1\text{ L}}{{{10}^{3}}\text{ }{{\text{m}}^{3}}} \right) \\ & =85\times {{10}^{-3}}\text{ g/}{{\text{m}}^{3}} \end{align} The mass of oxygen in each nanocontainer is calculated as follows: $m=\left( V \right)\left( d \right)$ Here V is volume and d is density. Substitute $85\times {{10}^{-3}}\text{ g/}{{\text{m}}^{\text{3}}}$ for d and $1.6\times {{10}^{-23}}\text{ }{{\text{m}}^{3}}$ for V. \begin{align} & m=\left( 1.6\times {{10}^{-23}}\text{ }{{\text{m}}^{3}} \right)\left( 85\times {{10}^{-3}}\text{ g/}{{\text{m}}^{\text{3}}} \right) \\ & =1.3\times {{10}^{-24}}\text{ g} \end{align} The mass of oxygen in each nanocontainer is $\underline{1.3\times {{10}^{-24}}\text{ g}}$. (c) The volume of oxygen inhaled by human per hour is calculated as follows: \begin{align} & V=\left( 0.50\text{ }\frac{\text{L}}{\text{breath}} \right)\left( 20\frac{\text{breaths}}{\text{min}} \right)\left( \text{60 }\frac{\min }{\text{hour}} \right) \\ & =600\text{ L/h} \end{align} The mass of oxygen inhaled is calculated as follows: $m=\left( V \right)\left( d \right)$ Here V is volume and d is density. Substitute $\text{0}\text{.28 g/L}$ for d and $600\text{ L}$ for V. \begin{align} & m=\left( 600\text{ L} \right)\left( \text{0}\text{.28 g/L} \right) \\ & =1.7\times {{10}^{2}}\text{ g} \end{align} The amount of oxygen inhaled by human per hour is $\underline{1.7\times {{10}^{2}}\text{ g}}$. (d) The minimum number of nanocontainers required to provide oxygen for one hour is equal to the ratio of the total mass of oxygen per hour to the mass of oxygen in one container as follows: \begin{align} & n=\frac{1.7\times {{10}^{2}}\text{ g}}{1.3\times {{10}^{-18}}\text{ g}} \\ & =1.3\times {{10}^{20}} \end{align} The minimum number of nanocontainers required to provide oxygen for one hour is $\underline{1.3\times {{10}^{20}}}$. (e) The minimum volume occupied by a number of nanocontainers is equal to the number of nanocontainers multiplied by the volume of each nanocontainer. \begin{align} & {{V}_{\text{min}}}=\left( 1.3\times {{10}^{20}} \right)\left( 1.6\times {{10}^{-23}}\text{ }{{\text{m}}^{3}} \right)\left( \frac{1\text{ L}}{{{10}^{-3}}\text{ }{{\text{m}}^{3}}} \right) \\ & =2.1\text{ L} \end{align} The minimum volume occupied by the number of nanocontainers is $\underline{2.1\text{ L}}$. The total volume of blood in an adult is around $5\text{ L}$. So, this is not feasible.