Some substances that are only very slightly soluble in water will spread over the surface of water to produce a film that is called a monolayer because it is only one molecule thick. A practical use of this phenomenon is to cover ponds to reduce the loss of water by evaporation. Stearic acid forms a monolayer on water. The molecules are arranged upright and in contact with one another, rather like pencils tightly packed and standing upright in a coffee mug. The model below represents an individual stearic acid molecule in the monolayer. (a) How many square meters of water surface would be covered by a monolayer made from \(10.0 \mathrm{g}\) of stearic acid? [Hint: What is the formula of stearic acid?] (b) If stearic acid has a density of \(0.85 \mathrm{g} / \mathrm{cm}^{3}\), estimate the length (in nanometers) of a stearic acid molecule. [Hint: What is the thickness of the monolayer described in part a?] (c) A very dilute solution of oleic acid in liquid pentane is prepared in the following way: $$\begin{aligned} &1.00 \mathrm{mL} \text { oleic acid }+9.00 \mathrm{mL} \text { pentane } \rightarrow \text { solution }(1)\\\ &1.00 \mathrm{mL} \text { solution }(1)+9.00 \mathrm{mL} \text { pentane } \rightarrow \text { solution }(2)\\\ &1.00 \mathrm{mL} \text { solution }(2)+9.00 \mathrm{mL} \text { pentane } \rightarrow \text { solution }(3)\\\ &1.00 \mathrm{mL} \text { solution }(3)+9.00 \mathrm{mL} \text { pentane } \rightarrow \text { solution }(4) \end{aligned}$$ A 0.10 mL sample of solution (4) is spread in a monolayer on water. The area covered by the monolayer is \(85 \mathrm{cm}^{2} .\) Assume that oleic acid molecules are arranged in the same way as described for stearic acid, and that the cross-sectional area of the molecule is \(4.6 \times 10^{-15} \mathrm{cm}^{2}\). The density of oleic acid is \(0.895 \mathrm{g} / \mathrm{mL} .\) Use these data to obtain an approximate value of Avogadro's number.

Step by step solution

01

Calculate the number of moles

The first step is to calculate the number of moles of stearic acid in the 10.0g and the 1.00mL of oleic acid. The molar mass of stearic acid is \(284.5 g/mol\), and the molar mass of oleic acid is \(282.5 g/mol\). To calculate the number of moles, we divide the given mass by the molar mass. Thus, there are \(10.0 g / 284.5 g/mol = 0.035 mol\) of stearic acid, and \(0.895 g / 282.5 g/mol = 0.00317 mol\) of oleic acid.

02

Calculate the area covered by the monolayer

Next, calculate how many molecules are in the monolayer. Avogadro's number tells us how many molecules are in one mole. Therefore, there are \(0.035mol * 6.022*10^{23} / mol = 2.1077*10^{22}\) stearic acid molecules. Since stearic acid molecules are standing upright, their cross sectional area is the area of the water surface each molecule can cover. The given cross-sectional area of stearic acid molecules is \(4.6*10^{-15} cm^2\). Thus, the total area that 10g of stearic acid can cover is \(2.1077*10^{22} * 4.6*10^{-15} cm^2 = 97 m^2\)

03

Calculate the length of a stearic acid molecule

To find the length of a stearic acid molecule, we need to convert the mass of stearic acid to a volume, by using the density formula. The volume of stearic acid is \(10.0g / 0.85 g/cm^3 = 11.76 cm^3\). Because stearic acid forms a monolayer, the volume of stearic acid is approximately equal to the area covered by the monolayer multiplied by the thickness of the monolayer: \(11.76 cm^3 = 97 m^2 * Thickness\). Solving for Thickness, we get \(Thickness = 11.76 cm^3 / 97 m^2 = 1.21 * 10^{-7}m = 1.21 nm\)

04

Calculate Avogadro's number

Finally, we can calculate Avogadro's number based on the information given for oleic acid. With 0.10mL of oleic acid solution spread in a monolayer covering 85 cm^2 area, we first calculate the number of molecules from the known cross-sectional area: \(85 cm^2 / 4.6*10^{-15} cm^2 = 1.843*10^{16}\) molecules. The moles of oleic acid in 0.10mL of solution can be determined from the dilution and the original volume, yielding 0.00317 moles of oleic acid. Avogadro's number is the number of molecules in one mole, therefore \(1.843*10^{16} molecules / 0.00317 mol = 5.81*10^{23} /mol\)

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