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 densityso $b_O={\rho }_b^{H_{2}O}/n_H-2b_H\simeq 5.81\times 10^{-15}\mathrm{m}$. Therefore $b_{D_{2}O}=2b_D+b_O\simeq 1.915\times 10^{-14}\mathrm{m}$ andThe assumptions are isotopically pure waters, additive nuclear scattering lengths, the same oxygen isotope contribution, and molar masses approximately $18$ and $20\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$. Equal molar volumes would give the rougher estimate $20/18=1.11\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}$.