The sedimentation coefficient of lysozyme $\left(\mathrm{M}=14,100 \mathrm{~g} \mathrm{~mol}^{-1}\right)$ in water at $20^{\circ} \mathrm{C}$ is $1.91 \times 10^{-13} \mathrm{~s}$ and the specific volume is $0.703 \mathrm{~cm}^3 \mathrm{~g}^{-1}$. The density of water at this temperature is $0.998 \mathrm{~g} \mathrm{~cm}^{-3}$ and $\eta=1.002 \mathrm{cP}$. Assuming lysozyme is spherical, what is the radius of this protein?
The frictional coefficient is dependent on molecular radius. Solving Equation (17.56) for $f$, we obtain
$$
\begin{aligned}
& f=\frac{m(1-\overline{\mathrm{V}} \rho)}{\overline{\mathrm{s}}}=\frac{\frac{\left(14,100 \mathrm{~g} \mathrm{~mol}^{-1}\right)}{\left(6.022 \times 10^{23} \mathrm{~mol}^{-1}\right)}\left(1-(0.703 \mathrm{~mL} \mathrm{~g})\left(0.998 \mathrm{~g} \mathrm{~mL}^{-1}\right)\right)}{1.91 \times 10^{-13} \mathrm{~s}} \\
& =3.66 \times 10^{-8} \mathrm{~g} \mathrm{~s}^{-1} \\
&
\end{aligned}
$$
The frictional coefficient is related to the radius for a spherical particle by Equation (17.51) such that
$$
r=\frac{f}{6 \pi \eta}=\frac{3.66 \times 10^{-8} \mathrm{~g} \mathrm{~s}^{-1}}{6 \pi\left(1.002 \mathrm{~g} \mathrm{~m}^{-1} \mathrm{~s}^{-1}\right)}=1.94 \times 10^{-9} \mathrm{~m}=1.94 \mathrm{~nm}
$$