 Solution

ID: past-exam-of-the-chemistry-course-of-the-year-3-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-4/64/b/i/solution

ρ_{b}^{S i} = \frac{2.33 \times 1 0 ^{6}}{28} (6.022 \times 1 0^{23}) (4.15 \times 1 0^{- 15}) = 2.080 \times 1 0^{14} m^{- 2} = 2.080 \times 1 0^{- 6} \overset{˚}{A}^{- 2} .

The density was converted to

g m^{- 3}

to match the molar mass in grams per mole, and

1 m^{- 2} = 1 0^{- 20} \overset{˚}{A}^{- 2}

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