## Chemistry: Molecular Approach (4th Edition)

Since $1\text{ lb}=453.59\text{ g}$, convert the mass of the person into grams as follows: \begin{align} & m=155\text{ lb}\left( \frac{453.59\text{ g}}{1\text{ lb}} \right) \\ & =7.03\times {{10}^{4}}\text{ g} \end{align} Now, the volume of the person is calculated as follows: $V=\frac{m}{d}$ Here, m is the mass and d is the density. Substitute $7.03\times {{10}^{4}}\text{ g}$ for m and $1.0\text{ g/c}{{\text{m}}^{\text{3}}}$ for d: \begin{align} & V=\frac{7.03\times {{10}^{4}}\text{ g}}{1.0\text{ g/c}{{\text{m}}^{\text{3}}}} \\ & =7.03\times \text{1}{{\text{0}}^{4}}\text{ c}{{\text{m}}^{3}} \end{align} Now, the height of the person is $4\text{ ft}$. One foot is equal to 30.48 cm. Thus, \begin{align} & h=4\text{ ft}\left( \frac{30.48\text{ cm}}{1\text{ ft}} \right) \\ & =121.92\text{ cm} \end{align} Now, the volume of the cylinder is as follows: $V=\pi {{r}^{2}}h$ Rearrange the equation: ${{r}^{2}}=\frac{V}{\pi h}$ Substitute $121.92\text{ cm}$ for h and $7.03\times \text{1}{{\text{0}}^{4}}\text{ c}{{\text{m}}^{3}}$ for V: \begin{align} & {{r}^{2}}=\frac{7.03\times \text{1}{{\text{0}}^{4}}\text{ c}{{\text{m}}^{3}}}{\left( 3.14 \right)\left( 121.92\text{ cm} \right)} \\ & r=\sqrt{\frac{7.03\times \text{1}{{\text{0}}^{4}}\text{ c}{{\text{m}}^{3}}}{\left( 3.14 \right)\left( 121.92\text{ cm} \right)}} \\ & =13.55\text{ cm} \end{align} The person is modeled by a cylinder. Thus, the waist of the person is equal to the circumference of the circle. Thus, the waist is calculated as follows: $w=2\pi r$ Substitute $13.55\text{ cm}$ for r: \begin{align} & {{w}_{1}}=2\left( 3.14 \right)\left( 13.55\text{ cm} \right) \\ & =85.1\text{ cm} \end{align} Now, calculate the volume of the fat as follows: \begin{align} & {{V}_{\text{f}}}=\left( \frac{40.0\text{ lb}}{0.918\text{ g/c}{{\text{m}}^{\text{3}}}} \right)\left( \frac{453.59\text{ g}}{1\text{ lb}} \right) \\ & =1.98\times {{10}^{4}}\text{ c}{{\text{m}}^{3}} \end{align} The total volume of the person is the sum of the volume of the person and the volume of the fat. Thus, \begin{align} & {{V}_{\text{T}}}=\left( 7.03\times {{10}^{4}}+1.98\times {{10}^{4}} \right)\text{ c}{{\text{m}}^{3}} \\ & =9.01\times {{10}^{4}}\text{ c}{{\text{m}}^{3}} \end{align} Now, the radius of cylinder is as follows: \begin{align} & {{r}^{2}}=\frac{9.01\times \text{1}{{\text{0}}^{4}}\text{ c}{{\text{m}}^{3}}}{\left( 3.14 \right)\left( 121.92\text{ cm} \right)} \\ & r=\sqrt{\frac{9.01\times \text{1}{{\text{0}}^{4}}\text{ c}{{\text{m}}^{3}}}{\left( 3.14 \right)\left( 121.92\text{ cm} \right)}} \\ & =15.34\text{ cm} \end{align} Now, as the waist of the person is equal to circumference of the circle with radius r, \begin{align} & {{w}_{2}}=2\left( 3.14 \right)\left( 15.34\text{ cm} \right) \\ & =96.34\text{ cm} \end{align} Now, the percentage increase in the waist size is calculated as follows: $\text{percentage}\,=\frac{{{w}_{2}}-{{w}_{1}}}{{{w}_{1}}}\times 100$ Substitute $96.34\text{ cm}$ for ${{w}_{2}}$ and $\text{85}\text{.1 cm}$ for ${{w}_{1}}$: \begin{align} & \text{percentage}\,=\frac{96.34\text{ cm}-85.1\text{ cm}}{\text{85}\text{.1 cm}}\times 100 \\ & =13\,\text{percent} \end{align} The percentage increase in the waist is 13.