 Solution

Interpret the stated free energy as per lattice site of volume $v_0$. Flory–Huggins solution theory counts polymer chains, giving the entropy term $(\phi /N)\ln \phi$, and solvent molecules, giving $(1-\phi )\ln (1-\phi )$. The interaction term penalizes unlike contacts when $\chi >0$; a simple contact model has $\chi =z[{ϵ}_{ps}-({ϵ}_{pp}+{ϵ}_{ss})/2]/(k_{B}T)$. Connectivity reduces polymer translational entropy relative to independent monomers. This is an incompressible mean-field model with composition-independent $\chi$ for the calculations below.^[ref]

Expanding the dilute osmotic pressure identifies the effective second virial coefficient. A good solvent has $\chi <1/2$ and positive excluded volume, promoting swollen chains. A poor solvent has $\chi >1/2$, favouring contraction and potentially phase separation. The theta-solvent condition is $\chi =1/2$ in this long-chain mean-field description: the second virial term vanishes, but higher terms remain. For finite $N$, this condition is not the exact critical point, which instead has ${\chi }_c=(1+N^{-1/2}{)}^2/2$.

Let $g=f_{\mathrm{mix}}/(k_{B}T)$. The osmotic pressure in Flory-Huggins solution theory isThus $\Pi =(k_{B}T/v_0)[\phi /N+(1/2-\chi ){\phi }^2+{\phi }^3/3+\cdots ]$. For mass concentration $c={\rho }_p\phi$ and chain molar mass $M_n=N{\rho }_p{v}_0{N}_A$,osmometry uses a membrane permeable to solvent but retaining every polymer chain, at known uniform temperature. Extrapolate $\Pi /c$ to zero concentration to remove interactions; the intercept gives the number-average molar mass. Aggregation, leakage of small chains, ions with a Donnan contribution, or a poor-solvent two-phase sample invalidate the simple molecular-weight interpretation.

The measured ratios give an exactly linear reduced-pressure fit:The reduced osmotic-pressure intercept is $185\mathrm{Pa}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}$. TODO: the osmometry temperature is not stated in Question 3, so its absolute numerical molar mass is undetermined. The data establish $M_n/T=0.04494\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$. If $T=300\mathrm{K}$ is additionally assumed, the conditional result is $M_n\simeq 13.5\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}=1.35\times 10^4\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$. temperatures stated in other questions are not silently imported.