Chapter 22, Problem 22.17P

(a)

Interpretation Introduction

Interpretation:

The volume of fluorine gas at given pressure and temperature has to b calculated.

Concept introduction:

Ideal gas Equation:

Any gas is described by using four terms namely pressure, volume, temperature and the amount of gas. Thus combining three laws namely Boyle’s, Charles’s Law and Avogadro’s Hypothesis the following equation could be obtained. It is referred as ideal gas equation.

  $\begin{array}{l}\text{V }\propto \frac{\text{nT}}{\text{P}}\\ \text{V = R}\frac{\text{nT}}{\text{P}}\\ \text{PV = nRT}\end{array}$

Where,

    $\begin{array}{l}\text{n = moles}\text{of}\text{gas}\\ \text{P = pressure}\\ \text{T = temperature}\\ \text{R = gas }\text{constant}\end{array}$

Explanation of Solution

The number of moles of fluorine and ${\text{SiF}}_{\text{4}}$ is calculated as,

$\begin{array}{l}\text{Moles of F =}\left(\text{100}{\text{. kg Ca}}_{\text{5}}{\text{ (PO}}_{\text{4}}{\text{)}}_{\text{3}}\text{F}\right)\left(\frac{{\text{10}}^{\text{3}}{\text{gCa}}_{\text{5}}{\text{ (PO}}_{\text{4}}{\text{)}}_{\text{3}}\text{F}}{\text{1kg}{\text{Ca}}_{\text{5}}{\text{ (PO}}_{\text{4}}{\text{)}}_{\text{3}}\text{F}}\right)\left(\frac{\text{1mol}{\text{Ca}}_{\text{5}}{\text{ (PO}}_{\text{4}}{\text{)}}_{\text{3}}\text{F}}{\text{504}\text{.31 g}{\text{Ca}}_{\text{5}}{\text{ (PO}}_{\text{4}}{\text{)}}_{\text{3}}\text{F}}\right)\left(\frac{\text{1mol}\text{F}}{\text{1mol}{\text{Ca}}_{\text{5}}{\text{ (PO}}_{\text{4}}{\text{)}}_{\text{3}}\text{F}}\right)\left(\frac{\text{15\%}}{\text{100\%}}\right)\\ \\ \text{=}\text{29}\text{.74361 mol F}\\ \\ {\text{Moles of SiF}}_{\text{4}}\text{ =}\left(\text{29}\text{.74361 mol F}\right)\left(\frac{\text{1mol}{\text{SiF}}_{\text{4}}}{\text{4mol}\text{F}}\right)\text{=}\text{7}{\text{.43590 mol SiF}}_{\text{4}}\end{array}$

The volume of fluorine gas is determined using ideal gas equation.

    $\begin{array}{l}\text{PV = nRT}\\ \\ \text{V}\text{=}\frac{\text{nRT}}{\text{P}}\text{=}\frac{\left(\text{7}{\text{.43590 mol SiF}}_{\text{4}}\right)\left(\text{0}\text{.0821}\frac{\text{L}\text{.atm}}{\text{mol}\text{.}\text{atm}}\right)\left(\left(\text{273 1450}\right)\text{K}\right)}{\text{1atm}}\\ \\ \text{=}\text{1}\text{.1}\text{×}{\text{10}}^{\text{3}}\text{L}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The volume of drinking water can be fluorinated to given level has to be calculated.

Concept introduction:

${\text{1 m}}^{\text{3}}\text{ = 1000L}$

Density of water = $\text{1}\text{.00 g/mL }\text{.}$

Explanation of Solution

The number of moles of ${\text{Na}}_{\text{2}}{\text{SiF}}_{\text{6}}$ and mass of fluorine are determined as,

    $\begin{array}{l}{\text{Moles of Na}}_{\text{2}}{\text{SiF}}_{\text{6}}\text{ =}\text{(7}{\text{.43590 mol SiF}}_{\text{4}}\text{)}\left(\frac{\text{1mol}{\text{Na}}_{\text{2}}{\text{SiF}}_{\text{6}}}{\text{2mol}{\text{SiF}}_{\text{4}}}\right)\text{=}\text{3}\text{.71795 mol}{\text{Na}}_{\text{2}}{\text{SiF}}_{\text{6}}\\ \\ {\text{Mass (g) of F}}^{\text{-}}\text{=}\left(\text{3}\text{.71795 mol}{\text{Na}}_{\text{2}}{\text{SiF}}_{\text{6}}\right)\left(\frac{{\text{6 mol F}}^{\text{-}}}{\text{1mol}{\text{Na}}_{\text{2}}{\text{SiF}}_{\text{6}}}\right)\left(\frac{\text{19g}{\text{F}}^{\text{-}}}{\text{1mol}{\text{F}}^{\text{-}}}\right)\text{=}\text{4}{\text{.2x10}}^{\text{2}}{\text{ g F}}^{\text{-}}\end{array}$

If one gram F– will fluoridate ${\text{10}}^{\text{6}}\text{ g}$ water to a level of one ppm, then it is to determine how many grams (converted to mL using density, then converted from ${\text{mL to L to m}}^{\text{3}}$) of water can $\text{423}{\text{.8463 g F}}^{\text{–}}$ fluoridate to the one ppm level.

Necessary conversion factors: ${\text{1 m}}^{\text{3}}\text{ = 1000L}$, density of water = $\text{1}\text{.00 g/mL }\text{.}$

$\text{Volume}\text{=}\left(\text{423}{\text{.8463 gF}}^{\text{-}}\right)\left(\frac{{\text{10}}^{\text{6}}\text{g}\text{water}}{\text{1g}{\text{F}}^{\text{-}}}\right)\left(\frac{\text{mL}\text{water}}{\text{1g}\text{water}}\right)\left(\frac{{\text{1cm}}^{\text{3}}}{\text{1mL}}\right){\left(\frac{{\text{10}}^{\text{-2}}\text{m}}{\text{1cm}}\right)}^{\text{3}}\text{=}\text{4}{\text{.2x10}}^{\text{2}}{\text{ m}}^{\text{3}}$