Number-average molecular weight from osmometry: 23-Chem-B8 Polymer Engineering, December 2014, Q2

Question 2: Number-average molecular weight from osmometry (Part A — 20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Osmotic pressures (as heights of a toluene column) versus concentration for PS in toluene at $T=378\ \text{K}$; $\rho_{\text{tol}}=785\ \text{kg/m}^3$, $R=8.3143\ \text{J mol}^{-1}\text{K}^{-1}$. A column height $h$ converts to pressure by $\Pi=\rho g h$; note $1\ \text{g/L}=1\ \text{kg/m}^3$.

Find. $M_n$ from the van’t Hoff/virial osmotic-pressure equation.

Reduced osmotic pressure $\Pi/c$ versus concentration. The intercept at $c\to0$ equals $RT/M_n$; the positive slope (second virial coefficient $A_2>0$) shows toluene is a good solvent.

Approach. Convert each height to a pressure, form the reduced osmotic pressure $\Pi/c$, extrapolate linearly to $c\to0$; the intercept equals $RT/M_n$.

Convert column heights to pressures. $\Pi=\rho g h$ with $\rho=785\ \text{kg/m}^3$, $g=9.81\ \text{m/s}^2$; each cm of toluene = $785\times9.81\times0.01 = 77.0\ \text{Pa}$. Thus $\Pi = 112.4,\,172.5,\,243.3,\,348.1,\,442.0,\,623.8\ \text{Pa}$.
Reduced osmotic pressure. With $c$ in kg/m³ (= g/L), $\Pi/c = 83.3,\,86.3,\,90.1,\,93.8,\,97.8,\,105.0\ \text{Pa}\,\text{m}^3\text{kg}^{-1}$. The virial form is $$\frac{\Pi}{c}=\frac{RT}{M_n}+RT\,A_2\,c+\cdots$$ so a plot of $\Pi/c$ vs.\ $c$ is linear with intercept $RT/M_n$.
Extrapolate to infinite dilution. Least squares gives slope $4.67$ and $$\left(\frac{\Pi}{c}\right)_{c\to0} = \boxed{77.0\ \text{Pa}\,\text{m}^3\text{kg}^{-1}}$$
Solve for $M_n$. $RT = 8.3143\times378 = 3142.8\ \text{J/mol}$, so $$M_n = \frac{RT}{(\Pi/c)_{0}} = \frac{3142.8}{77.0} = 40.8\ \text{kg/mol} = \boxed{4.08\times10^{4}\ \text{g/mol}}$$

Quantity	Value
Intercept $(\Pi/c)_{c\to0}=RT/M_n$	77.0 Pa·m³/kg
Second virial coefficient $A_2$	slope$/RT = 1.49\times10^{-3}\ \text{m}^3\text{mol kg}^{-2}$ ($>0$)
Number-average molecular weight $M_n$	≈ 4.08 × 10⁴ g/mol

Topic: Membrane osmometry — a colligative, absolute method for $M_n$.

Key relations: $\Pi=\rho g h$; virial expansion $\Pi/c = RT/M_n + RT A_2 c + \cdots$; intercept $=RT/M_n$, slope $=RT A_2$.

Why this works: Osmotic pressure counts moles of solute independent of their size, so the number-average (not weight-average) molecular weight results. The finite slope reflects polymer–solvent interactions; a positive $A_2$ means the solvent is thermodynamically good and coils repel, which is why the reduced pressure rises with concentration. Extrapolating removes these interactions and leaves the ideal $RT/M_n$ term.

Common pitfall: Forgetting to convert the manometric height (cm of solvent) into an SI pressure via $\rho g h$, or reading $M_n$ off a single dilute point instead of the intercept. Osmometry is blind to low-MW tails that leak through the membrane, so it can over-estimate $M_n$ for broad samples.

Source: Rudin & Choi 3rd ed., Ch. 3 (colligative properties); Painter & Coleman, Fundamentals of Polymer Science, Ch. on molecular weight.

Quantity	Value
Intercept \((\Pi/c)_{c\to0}=RT/M_n\)	77.0 Pa·m³/kg
Second virial coefficient \(A_2\)	slope\(/RT = 1.49\times10^{-3}\ \text{m}^3\text{mol kg}^{-2}\) (\(>0\))
Number-average molecular weight \(M_n\)	≈ 4.08 × 10⁴ g/mol

Question 2 of 7: Number-average molecular weight from osmometry

Question 2: Number-average molecular weight from osmometry (Part A — 20 marks)