## Chemistry: Molecular Approach (4th Edition)

$\underline{7.3\times {{10}^{11}}\text{ g/c}{{\text{m}}^{3}}}$
Convert the radius of the sun into meters as follows: \begin{align} & r=7.0\times {{10}^{5}}\text{ km}\left( \frac{{{10}^{3}}\text{ m}}{1\text{ km}} \right) \\ & =7.0\times {{10}^{8}}\ \text{cm} \end{align} Now, the volume of the sun is as follows: $V=\frac{4}{3}\pi {{r}^{3}}$ Substitute 3.14 for $\pi$ and $7.0\times {{10}^{10}}\ \text{cm}$ for r: \begin{align} & V=\frac{4}{3}\left( 3.14 \right){{\left( 7.0\times {{10}^{8}}\ \text{m} \right)}^{3}} \\ & =1.436\times {{10}^{27}}\text{ }{{\text{m}}^{3}} \end{align} Now, the mass of the sun is calculated as follows: $m=\left( d \right)\left( V \right)$ Here, d is the density and V is the volume. Substitute $1.4\times {{10}^{3}}\text{ kg/}{{\text{m}}^{3}}$ for d and $1.436\times {{10}^{27}}\text{ }{{\text{m}}^{3}}$ for V: \begin{align} & m=\left( 1.4\times {{10}^{3}}\text{ kg/}{{\text{m}}^{\text{3}}} \right)\left( 1.436\times {{10}^{27}}\text{ }{{\text{m}}^{3}} \right) \\ & =2.01\times {{10}^{30}}\text{ kg} \end{align} Now, the black hole has a mass of $1\times {{10}^{3}}\text{ }suns$. So, the mass of the black hole is calculated as follows: \begin{align} & {{m}_{\text{b}}}=\left( 2.01\times {{10}^{30}}\text{ kg} \right)\left( \frac{{{10}^{3}}\text{ g}}{1\text{ kg}} \right)\left( 1.0\times {{10}^{3}} \right) \\ & =2.01\times {{10}^{36}}\text{ g} \end{align} The diameter of the moon is $2.16\times {{10}^{3}}\text{ }miles$. Thus, the radius of the moon in centimeters will be as follows: \begin{align} & {{r}_{\text{m}}}=\left( \frac{2.16\times {{10}^{3}}\text{ miles}}{2} \right)\left( \frac{1.609\text{ km}}{1\text{ mile}} \right)\left( \frac{1000\text{ m}}{1\text{ km}} \right)\left( \frac{100\text{ cm}}{1\text{ m}} \right) \\ & =1.74\times {{10}^{8}}\text{ cm} \end{align} Now, the radius of the black hole is half the radius of the moon. Thus, the radius of the black hole is as follows: \begin{align} & {{r}_{\text{b}}}=\frac{1}{2}\left( 1.74\times {{10}^{8}}\text{ cm} \right) \\ & =8.69\times {{10}^{7}}\text{ cm} \end{align} So, the volume of the black hole is calculated as follows: \begin{align} & {{V}_{\text{b}}}=\frac{4}{3}\left( 3.14 \right){{\left( 8.69\times {{10}^{7}}\ \text{m} \right)}^{3}} \\ & =2.75\times {{10}^{24}}\text{ c}{{\text{m}}^{3}} \end{align} Now, the density of the black hole is calculated as follows: ${{d}_{\text{b}}}=\frac{{{m}_{\text{b}}}}{{{V}_{\text{b}}}}$ Substitute $2.75\times {{10}^{24}}\text{ c}{{\text{m}}^{3}}$ for ${{V}_{\text{b}}}$ and $2.01\times {{10}^{36}}\text{ g}$ for ${{m}_{\text{b}}}$: \begin{align} & {{d}_{\text{b}}}=\frac{2.01\times {{10}^{36}}\text{ g}}{2.75\times {{10}^{24}}\text{ c}{{\text{m}}^{3}}} \\ & =7.3\times {{10}^{11}}\text{ g/c}{{\text{m}}^{3}} \end{align} The density of the black hole is $\underline{7.3\times {{10}^{11}}\text{ g/c}{{\text{m}}^{\text{3}}}}$.