 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/ii/solution

Use the water data to determine the common oxygen contribution rather than assume identical molar volumes. Ordinary water has molecular number density

n_{H} = \frac{1 0 ^{6}}{18} N_{A} = 3.346 \times 1 0^{28} m^{- 3},

b_{O} = ρ_{b}^{H_{2} O} / n_{H} - 2 b_{H} ≃ 5.81 \times 1 0^{- 15} m

. Therefore

b_{D_{2} O} = 2 b_{D} + b_{O} ≃ 1.915 \times 1 0^{- 14} m

and

n_{D} = \frac{6.335 \times 1 0 ^{14}}{b _{D_{2} O}}, ρ_{D_{2} O} = \frac{20 n _{D}}{N _{A}} ≃ 1.10 g c m^{- 3} .

The assumptions are isotopically pure waters, additive nuclear scattering lengths, the same oxygen isotope contribution, and molar masses approximately

18

and

20 g mo l^{- 1}

. Equal molar volumes would give the rougher estimate

20/18 = 1.11 g c m^{- 3}

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