Avogadro's Number — Question 8

Section 19-1 · Physics pages 3046–3047

Question 8 A monolayer estimate links molecular count with surface area Use NA = 6.02214076 × 1023 mol−1, exact. A 1.00 mg sample has assigned molar mass 200 g/mol. Every molecule spreads into one complete layer with footprint .250 nm2, without overlap or uncovered gaps. Assume no material is lost. r one layer Schematic; labels, not drawing lengths, specify the data. (a) Calculate the covered area and the radius if the layer is circular. (b) If the observed area is only 60.0% of that prediction and the footprint is unchanged, infer the fraction of molecules in the monolayer. Can the area alone establish where the remaining molecules went?
Show solutionHide solution
Question 8 - Solution A monolayer estimate links molecular count with surface area r one layer (a) Method and calculation n = 10−3/200 = 5.00 × 10−6 mol, N = nNA = 3.01107038 × 1018. A = N(.250 × 10−18) = .752767595 m2, r = √ A/π ≈.489503 m. (b) Interpretation and further calculation flayer = Aobserved/Apredicted = .600. Nlayer = .600N, Nelsewhere = .400N. The area counts only molecules in the assumed single layer. Loss, dissolution or multilayer clustering cannot be distinguished without additional evidence. Physical check Convert square nanometres with 10−18, not 10−9.

Question 8

A monolayer estimate links molecular count with surface area

Use NA=6.02214076×1023mol−1N_A=6.02214076\times10^{23}\,\mathrm{mol^{-1}}, exact. A 1.00 mg sample has assigned molar mass 200 g/mol. Every molecule spreads into one complete layer with footprint .250nm2.250\,\mathrm{nm^2}, without overlap or uncovered gaps. Assume no material is lost.

Diagram shown in the original SVG worksheet.

  1. Calculate the covered area and the radius if the layer is circular.

  2. If the observed area is only 60.0% of that prediction and the footprint is unchanged, infer the fraction of molecules in the monolayer. Can the area alone establish where the remaining molecules went?

Question 8 - Solution

A monolayer estimate links molecular count with surface area

Diagram shown in the original SVG worksheet.

(a) Method and calculation

n=10−3/200=5.00×10−6mol,N=nNA=3.01107038×1018.n=10^{-3}/200=5.00\times10^{-6}\,\mathrm{mol},\quad N=nN_A=3.01107038\times10^{18}.A=N(.250×10−18)=.752767595m2,r=A/π≈.489503m.A=N(.250\times10^{-18})=.752767595\,\mathrm{m^2},\quad r=\sqrt{A/\pi}\approx.489503\,\mathrm m.

(b) Interpretation and further calculation

flayer=Aobserved/Apredicted=.600.f_{\mathrm{layer}}=A_{\mathrm{observed}}/A_{\mathrm{predicted}}=.600.Nlayer=.600N,Nelsewhere=.400N.N_{\mathrm{layer}}=.600N,\quad N_{\mathrm{elsewhere}}=.400N.The area counts only molecules in the assumed single layer. Loss, dissolution or multilayer clustering cannot be distinguished without additional evidence.

Physical check

Convert square nanometres with 10−1810^{-18}, not 10−910^{-9}.