Chapter 13, Problem 43P

(a)

To determine

To Find: The number of moles of water in all the Earth’s oceans.

Answer to Problem 43P

  $6.55\times {10}^{22}\text{moles}$

Given:

Average depth, $d=3\text{km}=3000\text{m}$

Formula used:

The volume of the oceans:

  $V=0.75\left(A_s\right)\times \Delta y$

Here, $A_s$ is the surface area and $\Delta y$ is the depth.

Density of sea water is ${\rho }_{sw}=1025\text{}{\text{kg/m}}^{\text{3}}$ .

Write the expression for the number of moles:

  $n=\frac{m}{M}$

Here, $m$ is the mass = and $M$ is the molar mass.

Calculation:

As compared to the Earth’s radius, the average depth is very small. So, the volume of the ocean is calculated as that of a spherical shell with surface area $\left(4\pi R_{\text{earth}}{}^2\right)$ and thickness $\Delta y$ .

  $\begin{array}{l}V=0.75\left(4\pi R_{\text{earth}}{}^2\right)\times \Delta y\\ =0.75\left(4\times 3.14\times {\left(6.38\times {10}^6\right)}^2\right)\times 3000\\ =1.15\times {10}^{18}{\text{m}}^3\end{array}$

The number of moles is calculated with the help of seawater density and molecular weight of water,

  $\begin{array}{c}n=\frac{m}{M}\\ =\frac{V\times \rho }{M}\\ =\left(1.15\times {10}^{18}{\text{m}}^{\text{3}}\right)\times \left(\frac{1025\text{kg}}{{\text{m}}^{\text{3}}}\right)\times \left(\frac{1\text{mol}}{18\times {10}^{-3}\text{kg}}\right)\\ =6.55\times {10}^{22}\text{moles}\end{array}$

Conclusion:

The number of moles is $6.55\times {10}^{22}\text{moles}$ .

Explanation of Solution

Given:

Average depth, $d=3\text{km}=3000\text{m}$

Formula used:

The volume of the oceans:

  $V=0.75\left(A_s\right)\times \Delta y$

Here, $A_s$ is the surface area and $\Delta y$ is the depth.

Density of sea water is ${\rho }_{sw}=1025\text{}{\text{kg/m}}^{\text{3}}$ .

Write the expression for the number of moles:

  $n=\frac{m}{M}$

Here, $m$ is the mass = and $M$ is the molar mass.

Calculation:

As compared to the Earth’s radius, the average depth is very small. So, the volume of the ocean is calculated as that of a spherical shell with surface area $\left(4\pi R_{\text{earth}}{}^2\right)$ and thickness $\Delta y$ .

  $\begin{array}{l}V=0.75\left(4\pi R_{\text{earth}}{}^2\right)\times \Delta y\\ =0.75\left(4\times 3.14\times {\left(6.38\times {10}^6\right)}^2\right)\times 3000\\ =1.15\times {10}^{18}{\text{m}}^3\end{array}$

The number of moles is calculated with the help of seawater density and molecular weight of water,

  $\begin{array}{c}n=\frac{m}{M}\\ =\frac{V\times \rho }{M}\\ =\left(1.15\times {10}^{18}{\text{m}}^{\text{3}}\right)\times \left(\frac{1025\text{kg}}{{\text{m}}^{\text{3}}}\right)\times \left(\frac{1\text{mol}}{18\times {10}^{-3}\text{kg}}\right)\\ =6.55\times {10}^{22}\text{moles}\end{array}$

Conclusion:

The number of moles is $6.55\times {10}^{22}\text{moles}$ .

(b)

To determine

To Find: The number of molecules of water in all the Earth’s oceans.

Answer to Problem 43P

The number of molecules are $4\times {10}^{46}$ .

Given info:

The number of moles of water in all the Earth’s oceans is $6.55\times {10}^{22}\text{moles}$ .

Formula used:

Write the expression for number of molecules

  $N=n\times N_A$

Here,

  $n$ is the number of moles present.

  $N_A$ is the Avogadro’s number.

Calculation:

Write the expression for number of molecules

  $\begin{array}{c}N=n\times N_A\\ =\left(\text{6}{\text{.55×10}}^{\text{22}}\text{moles}\right)\text{×}\left(\frac{\text{6}{\text{.022×10}}^{\text{23}}\text{molecules}}{\text{1}\text{mole}}\right)\\ =4\times {10}^{46}\text{molecules}\end{array}$

Conclusion:

The number of molecules is $4\times {10}^{46}\text{molecules}$ .

Explanation of Solution

Given info:

The number of moles of water in all the Earth’s oceans is $6.55\times {10}^{22}\text{moles}$ .

Formula used:

Write the expression for number of molecules

  $N=n\times N_A$

Here,

  $n$ is the number of moles present.

  $N_A$ is the Avogadro’s number.

Calculation:

Write the expression for number of molecules

  $\begin{array}{c}N=n\times N_A\\ =\left(\text{6}{\text{.55×10}}^{\text{22}}\text{moles}\right)\text{×}\left(\frac{\text{6}{\text{.022×10}}^{\text{23}}\text{molecules}}{\text{1}\text{mole}}\right)\\ =4\times {10}^{46}\text{molecules}\end{array}$

Conclusion:

The number of molecules is $4\times {10}^{46}\text{molecules}$ .