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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/iii/solution

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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 / iii / Solution by

Codex  0 2026-10-20

Assuming additive liquid volumes, the scattering length density is linear in D

_{2}

O volume fraction

ϕ_{D}

:

2.080 = - 0.560 (1 - ϕ_{D}) + 6.335 ϕ_{D}, ϕ_{D} = 0.3828,

where all scattering length densities are in

1 0^{- 6} \overset{˚}{A}^{- 2}

. The

100 g

of heavy water occupies about

90.99 c m^{3}

. The required ordinary-water volume is

V_{H} = V_{D} \frac{1 - ϕ _{D}}{ϕ _{D}} ≃ 146.7 c m^{3} .

Thus add approximately $147 g$ of H $_{2}$ O. Volume fraction, rather than mass fraction, enters this contrast-matching equation. Isotope exchange to form HDO does not change the total nuclear scattering-length balance in the additive-volume approximation.

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