A sample of styrene is polymerized with the help of a little azoisobutyronitrile as initiator. Hydrolysis of the resulting polymer releases \(6 \mathrm{~g}\) of ammonia from \(10 \mathrm{~kg}\) of polymer. If all the end groups are derived from the initiating radical ( \(\alpha\) cyanoisopropyl), what is the average number of styrene units in the chain?

Short Answer

Expert verified
The average number of styrene units in the chain is approximately 272.

Step by step solution

01

(Step 1: Calculate moles of ammonia released)

(Divide the released ammonia by its molar mass: \[ n_{NH_3} = \frac{6 \mathrm{~g}}{17 \mathrm{\frac{g}{mol}}} \])
02

(Step 2: Obtain the moles of the initiating radical)

(The initiating radical provides 1 mol of \(NH_3\) for each mole of initiating radical, so: \[ n_{initiator} = n_{NH_3} \])
03

(Step 3: Calculate total moles of styrene used)

(Divide the given styrene mass by its molar mass: \[ n_{styrene} = \frac{10 \mathrm{~kg}}{104 \mathrm{\frac{g}{mol}}} \times 1000 \mathrm{\frac{g}{kg}} \])
04

(Step 4: Find average number of styrene units per chain)

(Divide the moles of styrene by moles of initiating radical: \[ n_{avg} = \frac{n_{styrene}}{n_{initiator}} \]) Now, we will perform the calculations to find the average number of styrene units per chain:
05

(Step 1: Calculate moles of ammonia released)

(\[ n_{NH_3} = \frac{6 \mathrm{~g}}{17 \mathrm{\frac{g}{mol}}} = 0.353 \mathrm{~mol} \])
06

(Step 2: Obtain the moles of the initiating radical)

(\[ n_{initiator} = n_{NH_3} = 0.353 \mathrm{~mol} \])
07

(Step 3: Calculate total moles of styrene used)

(\[ n_{styrene} = \frac{10 \mathrm{~kg}}{104 \mathrm{\frac{g}{mol}}} \times 1000 \mathrm{\frac{g}{kg}} = 96.154 \mathrm{~mol} \])
08

(Step 4: Find average number of styrene units per chain)

(\[ n_{avg} = \frac{n_{styrene}}{n_{initiator}} = \frac{96.154 \mathrm{~mol}}{0.353 \mathrm{~mol}} = 272.3 \]) The average number of styrene units in the chain is approximately 272.

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