= Solution
Use the same standard pressure and <temperature>, the same bond length, and the same electronic and vibrational-ground-state approximations. The <translational partition function> scales as <mass> to the power $3/2$, giving the factor $(47/44)^{3/2}$. For the original isotopologue, $I=32m_uR^2$ and the <rotational symmetry number> is two. For the isotope-substituted molecule, the two oxygen ends are distinguishable, so the symmetry number is one. Hence
$$
\frac{q_{\rm rot,new}}{q_{\rm rot,old}}=2\frac{33.91}{32}.
$$
The change in standard molar <entropy> is
$$
\Delta S_m=R\left[\frac32\log\frac{47}{44}+\log\left(2\frac{33.91}{32}\right)\right]
=7.0678\ \mathrm{J\,mol^{-1}\,K^{-1}},
$$
so
$$
\boxed{S_{m,\rm new}^{\circ}\simeq218.91\ \mathrm{J\,mol^{-1}\,K^{-1}}}.
$$
This uses a convention omitting a separate nuclear-spin <entropy>. If nuclear-spin degeneracies are instead included explicitly, the spin-$1/2$ carbon-13 nucleus adds a further $R\log2$ relative to the spin-zero original isotopologue; that convention must be stated rather than silently mixed with the quoted <partition functions>.
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