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(A) Calculate the number of moles of surfactant: $n = \frac{M 	imes V \, (mL)}{1000} = \frac{10^{-3} 	imes 10}{1000} = 10^{-8} \, mol$.Calculate the

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$2.0 \, pm$

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(A) Calculate the number of moles of surfactant: $n = \frac{M 	imes V \, (mL)}{1000} = \frac{10^{-3} 	imes 10}{1000} = 10^{-8} \, mol$.Calculate the number of molecules: $N = n 	imes N_A = 10^{-8} 	imes 6 	imes 10^{23} = 6 	imes 10^{15} \, 	ext{molecules}$.Area covered by one molecule $(a^2)$: $a^2 = \frac{	ext{Total Area}}{	ext{Number of molecules}} = \frac{0.24 \, cm^2}{6 	imes 10^{15}} = 0.04 	imes 10^{-15} \, cm^2 = 4 	imes 10^{-17} \, cm^2$.Calculate edge length $(a)$: $a = \sqrt{4 	imes 10^{-17} \, cm^2} = 2 	imes 10^{-8.5} \, cm \approx 6.32 	imes 10^{-9} \, cm$. Re-evaluating based on standard approximation: If $N_A = 6 	imes 10^{23}$, $a^2 = \frac{0.24}{10^{-5} 	imes 6 	imes 10^{23}} = 4 	imes 10^{-20} \, cm^2$.$a = \sqrt{4 	imes 10^{-20}} = 2 	imes 10^{-10} \, cm = 2 	imes 10^{-12} \, m = 2 \, pm$.

$10 \, mL$ of $1 \, mM$ surfactant solution forms a monolayer covering $0.24 \, cm^2$ on a polar substrate. If the polar head is approximated as a cube, what is its edge length?

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