Question

assuming that the earths radius is 6378Km and that a molecule of a benzene may be treated as a disc of radius 300 pm , calculate the mass of benzene needed to create a chain of molecules around the equator?

Answer 1

Answer on Question#58590 – Chemistry – Organic chemistry

Question: assuming that the earth's radius is $6378\mathrm{Km}$ and that a molecule of a benzene may be treated as a disc of radius $300~\mathrm{pm}$, calculate the mass of benzene needed to create a chain of molecules around the equator?

Solution:

$$\begin{array}{l} r(\text{Earth}) = 6378~\mathrm{km} = 6378 \cdot 10^{-3}~\mathrm{m} \\ c(\text{Earth}) = 2\pi r(\text{Earth}) = 2 \cdot 3.14 \cdot 6378 \cdot 10^{-3}~\mathrm{m} = 40053.84 \cdot 10^{-3}~\mathrm{m} \\ r(\text{benzene}) = 300~\mathrm{pm} = 300 \cdot 10^{-12}~\mathrm{m} \\ d(\text{benzene}) = 2r(\text{benzene}) = 600~\mathrm{pm} = 600 \cdot 10^{-12}~\mathrm{m} \\ N(C_6H_6) = \frac{r(\text{Earth})}{r(\text{benzene})} = \frac{40053.84 \cdot 10^3~\mathrm{m}}{600 \cdot 10^{-12}~\mathrm{m}} = 66.75 \cdot 10^{15} \\ m(C_6H_6) = \frac{N(C_6H_6) \cdot M(C_6H_6)}{N_A} = \frac{66.75 \cdot 10^{15} \cdot 70 \cdot \frac{8}{\text{mol}}}{6.02 \cdot 10^{25}~\mathrm{mol}^{-1}} = 864.9 \cdot 10^{-8}~\mathrm{g} = 8.65 \cdot 10^{-6}~\mathrm{g} = 8.65\mu\mathrm{g} \\ \end{array}$$

Answer: $8.65\mu\mathrm{g}$

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