Question

Polymer 0.2 g in 100 mL benzene. Relative viscosity = 1.5. Mark–Houwink: $[\eta] = K M^a$ with $a=0.5$, $K=0.001$. Find molecular weight.

Hint:

Always compute concentration in g/dL for viscosity relations. Intrinsic viscosity links directly to molecular weight via the Mark–Houwink constants.

Answer 1

Correct Answer: 6.24

Step 1: Polymer concentration.
\[ C = \frac{0.2}{100} = 0.002 \ \text{g/mL} = 0.02 \ \text{g/dL} \]
Step 2: Calculate reduced viscosity.
\[ \eta_{red} = \frac{\eta_{rel} - 1}{C} = \frac{1.5 - 1}{0.02} = 25 \ \text{dL/g} \]
Step 3: Intrinsic viscosity approximation.
For dilute solution, $[\eta] \approx \eta_{red} = 25$ dL/g.
Step 4: Apply Mark–Houwink equation.
\[ [\eta] = K M^a \Rightarrow 25 = 0.001 . M^{0.5} \] \[ M^{0.5} = \frac{25}{0.001} = 25000 \Rightarrow M = (25000)^2 = 6.25 \times 10^8 \ \text{g/mol} \]

Step 5: Express in required format.
\[ 6.25 \times 10^8 = 0.0625 \times 10^{10} \ \text{g/mol} \] Rounded: $0.06 \times 10^{10}$ g/mol. Final Answer: \[ \boxed{0.06 \times 10^{10} \ \text{g/mol}} \]