10 mL of 1mM surfactant solution forms a monolayer covering 0.24 cm2 on a polar substrate. If the polar head is approximated as a cube, what is its edge length?

Engineering

Chemistry

Mole and Its Calculation

Adsorption

Question

10 mL of 1mM surfactant solution forms a monolayer covering 0.24 cm² on a polar substrate. If the polar head is approximated as a cube, what is its edge length?

2.0 nm

2.0 pm

0.1 nm

1.0 pm

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$Moles = \frac{MVml}{1000} = \frac{10^{- 3} \times 10}{1000} = 10^{- 5} mole$

10^–5 N_A molecules covering area = 0.24 cm²

^{$1 - - - - - - - - = \frac{0 .24}{10^{- 5} N_{A}} {cm}^{2}$}

^{$\frac{0 .24}{10^{- 5} \times 6 \times 10^{23}} = a^{2}$}

a² = 4 × 10^–20 cm²

a = 2 × 10^–10 cm

a = 2 × 10^–12 m

a = 2 pm