Osmotic pressure from the solvent chemical potential

A membrane passes water but not a nonvolatile solute. Pure liquid water at pressure $P$ is on one side; an ideal solution with water mole fraction $x_w=0.9900$ is on the other side at pressure $P+\Pi$. Both sides are at $T=300.0\ \mathrm{K}$. Treat water as incompressible with molar volume $\bar v_w=18.0\times10^{-6}\ \mathrm{m^3\,mol^{-1}}$, use $R=8.314\ \mathrm{J\,mol^{-1}K^{-1}}$, and neglect the pressure dependence of $\bar v_w$. For the solution use \[ \mu_w(T,P+\Pi,x_w)=\mu_w^*(T,P+\Pi)+RT\ln x_w, \] and for pure water use $\mu_w^*(T,P)$. Starting from equality of water chemical potentials: 1. Derive $\Pi=-(RT/\bar v_w)\ln x_w$, including its sign. 2. Evaluate the exact pressure in MPa. 3. Replace $-\ln x_w$ by $1-x_w$ and evaluate the dilute approximation. Compare the two values and state which is larger. 4. Explain why inserting the solute mole fraction directly into the logarithm is not the stated solvent-equilibrium calculation. Use natural logarithms and retain at least four significant figures in the comparison.

Put the solution-side and pure-side water chemical potentials equal before moving any term.