# Question 9b11e

Aug 29, 2016

Here's what I got.

#### Explanation:

For part (a), use the molar mass of pyridine to calculate how many moles you have in that sample

0.068 color(red)(cancel(color(black)("g"))) * ("1 mole C"_5"H"_5"N")/(79.10 color(red)(cancel(color(black)("g")))) = 8.597 * 10^(-4)"moles C"_5"H"_5"N"

As you know, the number of particles needed to have one mole is given by Avogadro's number

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{1 mole" = 6.022 * 10^(23)"particles} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Your sample of pyridine will thus contain

8.597 * 10^(-4) color(red)(cancel(color(black)("moles C"_5"H"_5"N"))) * (6.022 * 10^(23)"molec C"_5"H"_5"N")/(1color(red)(cancel(color(black)("mole C"_5"H"_5"N"))))

$= \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{5.2 \cdot {10}^{20} \text{molec CH"_5"H"_5"N}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

For part (b), use the same approach to calculate the number of formula units present in your sample of zinc oxide.

You will have

5.0 color(red)(cancel(color(black)("g"))) * "1 mole ZnO"/(81.41color(red)(cancel(color(black)("g")))) = 6.142 * 10^(-2)"moles ZnO"

and

6.142 * 10^(-2) color(red)(cancel(color(black)("moles ZnO"))) * (6.022 * 10^(23)"f. units ZnO")/(1color(red)(cancel(color(black)("mole ZnO"))))

$= 3.7 \cdot {10}^{22} \text{f. units ZnO}$

The ratio between the number of molecules of pyridine and the number of formula units of zinc oxide will thus be

$\frac{5.2 \cdot {10}^{20}}{3.7 \cdot {10}^{22}} = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{0.014} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$\textcolor{w h i t e}{a}$

SIDE NOTE Notice that you can also calculate this ratio by dividing the number of moles of each compound

$\left(8.597 \cdot {10}^{- 4} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles C"_5"H"_5"N"))))/(6.142 * 10^(-2)color(red)(cancel(color(black)("moles ZnO}}}}\right) = 0.014$

That is the case because a mole is simply a fixed number of molecules / formula units.

$\textcolor{w h i t e}{a}$

For part (c), you know that zinc oxide has a surface area per gram equal to ${\text{42 m"^2"g}}^{- 1}$, which basically means that you can distribute $\text{1 g}$ of zinc oxide over a surface area of ${\text{42 m}}^{2}$.

Now, calculate how many molecules of pyridine are adsorbed per gram of zinc oxide by using the fact that you know how many molecules are adsorbed by $\text{5.0 g}$ of zinc oxide

1 color(red)(cancel(color(black)("g ZnO"))) * (5.2 * 10^(20)"molec. C"_5"H"_5"N")/(5.0 color(red)(cancel(color(black)("g ZnO"))))

$= 1.04 \cdot {10}^{20} \text{molec C"_5"H"_5"N}$

So, $\text{1 g}$ of zinc oxide adsorbs $1.04 \cdot {10}^{20}$ molecules of pyridine, it follows that the surface area per molecule of pyridine will be

48"m"^2 / color(red)(cancel(color(black)("1 g ZnO"))) * color(red)(cancel(color(black)("1 g ZnO")))/(1.04 * 10^(20)"molec. C"_5"H"_5"N")

$= \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{4.6 \cdot {10}^{- 19} \text{m"^2" / molec C"_5"H"_5"N}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answers are rounded to two sig figs.

SIDE NOTE If you want, you can convert this value from square meters per molecule to something like square nanometers per molecule

4.6 * 10^(-19) color(red)(cancel(color(black)("m"^2)))/("1 molec. C"_5"H"_5"N") * (10^9 * 10^9"nm"^2)/(1color(red)(cancel(color(black)("m"^2))))#

$= \text{0.46 nm"^2" / molec. C"_5"H"_5"N}$