Chapter F, Problem E.31E

(a)

Interpretation Introduction

Interpretation:

Amount of water that is brought and the money that is paid per liter of water has to be calculated.

Answer to Problem E.31E

The quantity of water brought is $\text{1}\text{.6}\text{kg}$ and the price that is paid per liter of water is $\$65.5$.

Explanation of Solution

Mass of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\cdot {\text{10H}}_{\text{2}}\text{O}$ is given as $\text{2}\text{.5}\text{kg}$.  The molar mass of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\cdot {\text{10H}}_{\text{2}}\text{O}$ is calculated as shown below;

$\begin{array}{c}\text{Molar}\text{mass}=2\times \text{M}(\text{Na})+1\times M(\text{C})+3\times M(\text{O})+10\times \text{M}({\text{H}}_{\text{2}}\text{O})\\ =2(22.99\text{g}\cdot {\text{mol}}^{-1})+1(12.01\text{g}\cdot {\text{mol}}^{-1})+3(16.00\text{g}\cdot {\text{mol}}^{-1})+10(18.016\text{g}\cdot {\text{mol}}^{-1})\\ =45.98\text{g}\cdot {\text{mol}}^{-1}+12.01\text{g}\cdot {\text{mol}}^{-1}+48.00\text{g}\cdot {\text{mol}}^{-1}+180.16\text{g}\cdot {\text{mol}}^{-1}\\ =286.16\text{g}\cdot {\text{mol}}^{-1}\end{array}$

The number of moles of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\cdot {\text{10H}}_{\text{2}}\text{O}$ in $\text{2}\text{.5}\text{kg}$ is calculated as shown below;

    $\begin{array}{c}{n}_{({\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\cdot {\text{10H}}_{\text{2}}\text{O})}=\frac{2500\text{g}}{286.16\text{g}\cdot {\text{mol}}^{-1}}\\ =8.74\text{mol}\end{array}$

One mole of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\cdot {\text{10H}}_{\text{2}}\text{O}$ contains ten moles of water.  Therefore, the number of moles of water in $\text{8}\text{.74}\text{mol}$ of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\cdot {\text{10H}}_{\text{2}}\text{O}$ is calculated as shown below;

    $\begin{array}{c}{n}_{({\text{H}}_{\text{2}}\text{O})}=8.74{\text{mol Na}}_{\text{2}}{\text{CO}}_{\text{3}}\cdot {\text{10H}}_{\text{2}}\text{O}\times \frac{10\text{mol}{\text{H}}_{\text{2}}\text{O}}{1{\text{mol Na}}_{\text{2}}{\text{CO}}_{\text{3}}\cdot {\text{10H}}_{\text{2}}\text{O}}\\ =87.4\text{mol}\end{array}$

Therefore, the mass of water present in $\text{2}\text{.5}\text{kg}$ is calculated as shown below;

    $\begin{array}{c}{m}_{({\text{H}}_{\text{2}}\text{O})}=87.4\text{mol}\times \frac{18.016\text{g}}{1\text{mol}}\\ =1574.59\text{g}\\ \text{=1}\text{.6}\text{kg}\end{array}$

Thus the mass of water that is present is $\text{1}\text{.6}\text{kg}$.  The mass of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ is calculated as shown below;

    $\begin{array}{c}\text{Mass}\text{of}{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}=\text{Total}{\text{mass of Na}}_{\text{2}}{\text{CO}}_{\text{3}}\cdot {\text{10H}}_{\text{2}}\text{O}-\text{Mass}\text{of}\text{water}\\ =2.5\text{kg}-1.6\text{kg}\\ =0.9\text{kg}\end{array}$

The cost of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$  with mass of $\text{2}\text{.5}\text{kg}$ is given as $\$195$.  Therefore, the price of $\text{0}\text{.9}\text{kg}$ of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ is calculated as follows;

    $\begin{array}{c}\text{Cost}\text{of}\text{0}\text{.9}\text{kg}{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}=0.9\text{kg}{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\times \frac{\$195}{2.5\text{kg}{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}}\\ =\$70.2\end{array}$

Cost of water:

The cost of water can be calculated by subtracting the price of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ from ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\cdot {\text{10H}}_{\text{2}}\text{O}$.

    $\begin{array}{c}\text{Cost}\text{of}{\text{H}}_{\text{2}}\text{O}=\$175-\$70.2\\ =\$104.8\end{array}$

The quantity of water obtained is $\text{1}\text{.6}\text{kg}$.  Therefore, the price per liter of water can be calculated as follows;

    $\begin{array}{c}\text{Price}\text{per}\text{liter}\text{of}\text{water}=\frac{\$104.8}{1.6\text{L}}\\ =\$65.5\end{array}$

Hence, the quantity of water brought is $\text{1}\text{.6}\text{kg}$ and the price that is paid per liter of water is $\$65.5$.

(b)

Interpretation Introduction

Interpretation:

If the water is valued at zero cost, then the fair price of the hydrated compound, ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\cdot {\text{10H}}_{\text{2}}\text{O}$ has to be given.

Answer to Problem E.31E

Fair price of the hydrated compound will be $\$70.2$.

Explanation of Solution

The mass of water that is present is $\text{1}\text{.6}\text{kg}$.  The mass of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ is calculated as shown below;

    $\begin{array}{c}\text{Mass}\text{of}{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}=\text{Total}{\text{mass of Na}}_{\text{2}}{\text{CO}}_{\text{3}}\cdot {\text{10H}}_{\text{2}}\text{O}-\text{Mass}\text{of}\text{water}\\ =2.5\text{kg}-1.6\text{kg}\\ =0.9\text{kg}\end{array}$

The cost of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$  with mass of $\text{2}\text{.5}\text{kg}$ is given as $\$195$.  Therefore, the fair price of $\text{0}\text{.9}\text{kg}$ of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ is calculated as follows;

    $\begin{array}{c}\text{Fair price}=0.9\text{kg}{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\times \frac{\$195}{2.5\text{kg}{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}}\\ =\$70.2\end{array}$

Hence, the fair price of hydrated compound is calculated as $\$70.2$.