A given sample of xenon fluoride contains molecules of a single type, XeF_n, where n is some whole number. Given that 9.03 * 10²⁰ molecules of XeF_n weigh 0.311 g, calculate n.

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The molecular weight of XeF_n can be calculated as follows: 1 mole of Xe = 131.29 g 1 mole of F = 18.99 g 1 mole of S = 32.06 g 1 mole of U = 238.03 g 1 mole of P = 30.97 g 1 mole of B = 10.81 g Molecular weight of XeF_n = 131.29 + 18.99 + 32.06 + 238.03 + (30.97 x 2) + (10.81 x n) = 452.96 + (10.81n) 0.311 g of XeF_n contains 9.03 × 10^10 molecules. The weight of one molecule is: 0.311 g ÷ (9.03 × 10^10) = 3.44 × 10^-12 g/molecule Therefore, the weight of (1 + n) atoms in one molecule is: 3.44 × 10^-12 x (1 + n) = 6.63 × 10^-23 g/atom To calculate n, we need to find out what value of n gives a total weight closest to the atomic weight of xenon, which is 131.29 g/mol. Let's test n = 2: Weight of XeF_n = 452.96 + (10.81 x 2) = 474.58 g/mol Weight of (1 + n) atoms in one molecule = 3.44 × 10^-12 x 3 = 1.03 × 10^-11 g/atom Total weight of (1 + n) atoms in 9.03 × 10^10 molecules = 9.31 × 10^-13 g Number of moles of XeF_n in 0.311 g = 0.311 g ÷ 474.58 g/mol = 6.55 × 10^-4 mol Number of atoms of (1 + n) in 6.55 × 10^-4 mol = 9.03 × 10^10 x (1 + n) Total weight of (1 + n) atoms in 6.55 × 10^-4 mol = (9.03 × 10^10 x (1 + n)) x (1.03 × 10^-11) = 9.31 × 10^-4 (1 + n) g Ratio of total weight of (1 + n) atoms to molecular weight of XeF_n = (9.31 × 10^-4 (1 + n) g) ÷ (0.311 g) = 2.98 × 10^-3 (1 + n) Let's test n = 4: Weight of XeF_n = 452.96 + (10.81 x 4) = 495.4 g/mol Weight of (1 + n) atoms in one molecule = 3.44 × 10^-12 x 5 = 1.72 × 10^-11 g/atom Total weight of (1 + n) atoms in 9.03 × 10^10 molecules = 1.55 × 10^-12 g Number of moles of XeF_n in 0.311 g = 0.311 g ÷ 495.4 g/mol = 6.27 × 10^-4 mol Number of atoms of (1 + n) in 6.27 × 10^-4 mol = 9.03 × 10^10 x (1 + n) Total weight of (1 + n) atoms in 6.27 × 10^-4 mol = (9.03 × 10^10 x (1 + n)) x (1.72 × 10^-11) = 1.23 × 10^-3 (1 + n) g Ratio of total weight of (1 + n) atoms to molecular weight of XeF_n = (1.23 × 10^-3 (1 + n) g) ÷ (0.311 g) = 3.96 × 10^-3 (1 + n) As the ratio of total weight of (1 + n) atoms to molecular weight of XeF_n is higher when n = 4 than when n = 2, we can conclude that n = 4 is the correct answer.

A Given Sample of Xenon Fluoride Contains Molecules of a Single