## Chemistry: Molecular Approach (4th Edition)

Published by Pearson

# Chapter 1 - Exercises - Page 43: 155

#### Answer

(a)Ice is less dense than water, which makes it float on water. The high heat capacity of water and an insulating floating sheet of ice above the water in ponds and lakes help fish to survive the winters. (b) (i)Pluto is mainly composed of ice, and not rocks. (ii)From the extrapolated data, the density of water on Pluto is higher than in Antarctica. (c) $\underline{\text{2}\text{.44}\times {{10}^{7}}\text{ k}{{\text{m}}^{3}}}$ (d)$\underline{\text{3}\text{.81}\times {{10}^{9}}\text{ k}{{\text{m}}^{3}}}$ Pluto has more water than Antarctica.

#### Work Step by Step

(a) The density of ice is ${{d}_{\text{ice}}}(0{}^\circ \text{C})=\text{918 kg/}{{\text{m}}^{3}}$ and density of water is${{d}_{\text{water}}}(4{}^\circ \text{C})=\text{999}\text{.97 kg/}{{\text{m}}^{3}}$. As the density of ice is less than that of water, it floats on the water. The specific heat of water is high, which makes the water in the ponds and lakes less colder than the air in winters. The surface water that is in direct contact of the cold air gets frozen, and because of its less density, floats on the water below it. This sheet of ice above the water acts as an insulator and helps the water in the lakes to stay warm, keeping the fish alive. (b) (i) Convert the mean density of Pluto as follows: \begin{align} & {{d}_{\text{Pluto}}}=\text{1}\text{.8}-\text{2}\text{.1 g/c}{{\text{m}}^{3}} \\ & =\text{1}\text{.8 g/c}{{\text{m}}^{3}}\times \left( \frac{1000\text{ kg/}{{\text{m}}^{3}}}{1\text{ g/c}{{\text{m}}^{3}}} \right)-\text{2}\text{.1 g/c}{{\text{m}}^{3}}\times \left( \frac{1000\text{ kg/}{{\text{m}}^{3}}}{1\text{ g/c}{{\text{m}}^{3}}} \right) \\ & =1800-2100\text{ kg/}{{\text{m}}^{3}} \end{align} The mean density of Earth’s moon is ${{d}_{\text{moon}}}\text{= 3346}\text{.4 kg/}{{\text{m}}^{3}}$. It is evident that the mean density of Pluto is much less than the mean density of Earth’s moon. The higher mean density of the Earth’s moon is due to the presence of rocks made up of oxygen, silicon, magnesium, iron, calcium, and aluminum. Therefore, the small mean density of Pluto indicates that it is mainly composed of ice, and not rocks. (ii) The temperatures on Pluto and Antarctica are $-\text{233}{}^\circ \text{C to }-\text{223}{}^\circ \text{C and }-\text{65}\text{.1}{}^\circ \text{C}$, respectively. The values for the density of water at these temperatures are calculated using FORECAST in Excel by taking the given values in the graph. ${{d}_{\text{water}}}(-65.1{}^\circ \text{C})=\text{925}\text{.7 kg/}{{\text{m}}^{3}}$ ${{d}_{\text{water}}}(-223{}^\circ \text{C})=\text{942}\text{.4 kg/}{{\text{m}}^{3}}$ ${{d}_{\text{water}}}(-233{}^\circ \text{C})=\text{943}\text{.5 kg/}{{\text{m}}^{3}}$ It is evident that the density of water on Pluto is higher than in Antarctica. (c) The value for the density of water at $-20{}^\circ \text{C}$ is calculated using FORECAST in Excel by taking the given values in the graph. ${{d}_{\text{water}}}(-20{}^\circ \text{C})=\text{920}\text{.92 kg/}{{\text{m}}^{3}}$ Calculate the total mass of ice as follows: \begin{align} & {{m}_{\text{ice}}}={{V}_{\text{ice}}}(-20{}^\circ \text{C})\times {{d}_{\text{water}}}(-20{}^\circ \text{C})\text{ } \\ & \text{=26}\text{.5 million k}{{\text{m}}^{3}}\times \left( \frac{1\times {{10}^{6}}}{1\text{ million}} \right)\times \left( \frac{1\times {{10}^{9}}{{\text{m}}^{3}}}{1\text{ k}{{\text{m}}^{3}}} \right)\times \text{920}\text{.92 kg/}{{\text{m}}^{3}} \\ & =\text{2}\text{.44}\times {{10}^{19}}\text{ kg} \end{align} ${{d}_{\text{water}}}(4{}^\circ \text{C})=\text{999}\text{.97 kg/}{{\text{m}}^{3}}$ Calculate the volume of water at $4{}^\circ \text{C}$ as follows: \begin{align} & {{V}_{\text{water}}}(4{}^\circ \text{C})=\frac{{{m}_{\text{water}}}}{{{d}_{\text{water}}}(4{}^\circ \text{C})} \\ & =\frac{\text{2}\text{.44}\times {{10}^{19}}\text{ kg}}{\text{999}\text{.97 kg/}{{\text{m}}^{3}}}\times \left( \frac{1\text{ k}{{\text{m}}^{3}}}{1\times {{10}^{9}}{{\text{m}}^{3}}} \right) \\ & =\text{2}\text{.44}\times {{10}^{7}}\text{ k}{{\text{m}}^{3}} \end{align} The volume of liquid water at $4{}^\circ \text{C}$ is $\underline{\text{2}\text{.44}\times {{10}^{7}}\text{ k}{{\text{m}}^{3}}}$. (d) Calculate the mass of ice on Pluto as follows: \begin{align} & {{m}_{\text{ice}}}(-175{}^\circ \text{C})=\frac{{{m}_{\text{Pluto}}}\times \text{mass percent of Ice}}{100}\text{ } \\ & =\frac{\text{1}\text{.27}\times \text{1}{{\text{0}}^{22}}\text{ kg}\times 30}{100} \\ & =\text{38}\text{.1}\times \text{1}{{\text{0}}^{20}}\text{ kg} \end{align} ${{d}_{\text{water}}}(4{}^\circ \text{C})=\text{999}\text{.97 kg/}{{\text{m}}^{3}}$ Calculate the volume of water at $4{}^\circ \text{C}$ as follows: \begin{align} & {{V}_{\text{water}}}(4{}^\circ \text{C})=\frac{{{m}_{\text{water}}}}{{{d}_{\text{water}}}(4{}^\circ \text{C})} \\ & =\frac{\text{38}\text{.1}\times \text{1}{{\text{0}}^{20}}\text{ kg}}{\text{999}\text{.97 kg/}{{\text{m}}^{3}}}\times \left( \frac{1\text{ k}{{\text{m}}^{3}}}{1\times {{10}^{9}}{{\text{m}}^{3}}} \right) \\ & =\text{3}\text{.81}\times {{10}^{9}}\text{ k}{{\text{m}}^{3}} \end{align} The volume of liquid water at $4{}^\circ \text{C}$ on Antarctica is $\text{2}\text{.44}\times {{10}^{7}}\text{ k}{{\text{m}}^{3}}$. Therefore, Pluto has more water than Antarctica. The volume of liquid water at $4{}^\circ \text{C}$ is $\underline{\text{3}\text{.81}\times {{10}^{9}}\text{ k}{{\text{m}}^{3}}}$ and Pluto has more water than Antarctica.

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