Question

The intrinsic viscosity of a sample of polystyrene in toluene is \(84\ \text{cm}^3\ \text{g}^{-1}\) at \(30\^{\circ}\text{C}\). It follows the Mark–Houwink equation with empirical constants \(K = 1.05\times 10^{-2}\ \text{cm}^3\ \text{g}^{-1}\) and \(a = 0.75\). The molecular weight of the polymer is ______\(\times10^{3}\ \text{g mol}^{-1}\) (rounded off to the nearest integer). 
 

Hint:

For Mark–Houwink, if \(1/a=\tfrac{4}{3}\), rewrite \(M=(x)^{4/3}=(x^{1/3})^{4}\) to compute quickly by taking a cube root first and then raising to the 4th power.

Answer 1

Correct Answer: 158

Step 1: Use the Mark–Houwink relation \([\,\eta\,] = K\,M^{a}\). Given \([\,\eta\,]=84\), \(K=1.05\times10^{-2}\), \(a=0.75\).
Step 2: Solve for \(M\): \[ M=\left(\frac{[\,\eta\,]}{K}\right)^{1/a} =\left(\frac{84}{1.05\times10^{-2}}\right)^{\!1/0.75} =\left(8000\right)^{4/3}. \] Step 3: Compute: \[ 8000^{1/3}=20 \;\Rightarrow\; M=20^{4}=160\,000\ \text{g mol}^{-1}=160\times10^{3}\ \text{g mol}^{-1}. \] Step 4: Rounded to the nearest integer for the \(\times10^{3}\) form gives \(\boxed{160}\).