Question

If 5 g of a monodisperse polystyrene sample of molecular weight 10,000 g mol$^{-1}$ is mixed with 15 g of another monodisperse polystyrene sample of molecular weight 20,000 g mol$^{-1}$, then the polydispersity of the resulting mixture is _________ (rounded off to two decimal places).

Hint:

The PDI is calculated by taking the ratio of \( M_w \) to \( M_n \), giving an indication of the spread in molecular weights.

Answer 1

Correct Answer: 1.04

The polydispersity index (PDI) is defined as the ratio of the weight average molecular weight (\( M_w \)) to the number average molecular weight (\( M_n \)): \[ PDI = \frac{M_w}{M_n} \] First, calculate \( M_n \) (number average molecular weight) for the mixture: \[ M_n = \frac{\sum m_i N_i}{\sum N_i} \] where \( m_i \) is the molecular weight and \( N_i \) is the number of moles for each sample. For the first sample: \[ N_1 = \frac{5}{10000} = 0.0005\ \text{mol} \] For the second sample: \[ N_2 = \frac{15}{20000} = 0.00075\ \text{mol} \] Now, calculate \( M_n \): \[ M_n = \frac{(10000)(0.0005) + (20000)(0.00075)}{0.0005 + 0.00075} \] \[ M_n = \frac{5 + 15}{1.25 \times 10^{-3}} = \frac{20}{1.25 \times 10^{-3}} = 16000\ \text{g/mol} \] Now, calculate \( M_w \) (weight average molecular weight) for the mixture: \[ M_w = \frac{\sum m_i^2 N_i}{\sum m_i N_i} \] \[ M_w = \frac{(10000)^2(0.0005) + (20000)^2(0.00075)}{(10000)(0.0005) + (20000)(0.00075)} \] \[ M_w = \frac{50000000 + 300000000}{20} = \frac{350000000}{20} = 17500000\ \text{g/mol} \] Finally, calculate the polydispersity: \[ PDI = \frac{M_w}{M_n} = \frac{17500000}{16000} = 1.09375 \] \[ \boxed{1.09} \]