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A sample of polystyrene prepared by heating styrene with tribromo benzoyl peroxide in the absence of air has the formula $B{r_3}{C_6}{H_3}{\left( {{C_8}{H_8}} \right)_n}$. The number of $n$ varies with the condition of preparation. One sample of polystyrene prepared in this manner was found to contain 10.46% of bromine. What is the value of $n$?

Answer
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Hint: We can use the idea of % of elements present in a compound to find the value of$n$. Though the question may seem complex but the solution is pretty simple. Since the percentage of bromine is given we can find the value of $n$ from that.

Complete Step by step answer:
Here, we are asked to find the value of $n$ in a polystyrene. We know that polymers contain large molecules composed of many repeating units. So, we needed to find the number of repeated units formed when polystyrene is prepared by heating styrene with tribromo benzoyl peroxide in the absence of air.
We have given that the formula of polystyrene formed is $B{r_3}{C_6}{H_3}{\left( {{C_8}{H_8}} \right)_n}$.
Also, it is said the number of $n$varies with the condition of preparation. Apart from this we know that one sample of polystyrene prepared in this manner was found to contain 10.46% of bromine.
Let us understand what that means. So, percentage of bromine means the percentage weight of bromine in that polystyrene. From the formula we know there are 3 Bromine atoms present.
$Molecular\;weight\;of\;Br = 80$
$\therefore Total\;weight\;of\;Br\;in\;polystyrene = 3 \times 80$
So we can say that,
$\% \;of\;Br\; = \dfrac{{\;weight\;of\;Br}}{{total\;weight\;of\;polystyrene}} \times 100$
Total weight of polystyrene is $3 \times 80 + 6 \times 12 + \left( {8 \times 12 + 8 \times 1} \right)n$
Where $Molecular\;weight\;of\;C = 12$
$Molecular\;weight\;of\;H = 1$
Therefore,
$10.46\% = \dfrac{{3 \times 80}}{{3 \times 80 + 6 \times 12 + \left( {8 \times 12 + 8 \times 1} \right)n}} \times 100$
$\Rightarrow 10.46 = \dfrac{{240}}{{332 + 104n}} \times 100$
On solving for $n$, we get
$10.87n = 240 - 32.87$
$\Rightarrow n = 19.05$

Therefore on approximation we can say $n$ is 90.

Note: Here the reaction involved has not much role to play. It is just some basic concepts of chemistry. We may get diverted by the reaction given in the question therefore make sure what is the question exactly and try to solve it.