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Chapter 8: Problem 66

A polystyrene of formula \(\mathrm{Br}_{3} \mathrm{C}_{6} \mathrm{H}_{2}\left(\mathrm{C}_{\mathrm{s}} \mathrm{H}_{\mathrm{s}}\right)_{n}\) was prepared by heating styrene with tribromobenzyl peroxide in the absence of air. It was found to contain \(10.46 \%\) bromine, by mass. The value of \(n\) is \((\mathrm{Br}=80)\) (a) 20 (b) 21 (c) 19 (d) 22

Short Answer

Expert verified
Answer: 21.

Step by step solution

01

Calculate the molar mass of the polystyrene and styrene monomer

To start the process, we need to find the molar mass of the polystyrene and the styrene monomer. The molar mass of styrene (C_sH_s) can be calculated as: M(C_sH_s)= M(C)+M(H) = 12*8 +1*8 = 96+8 =104 The given formula for polystyrene is Br3C6H2(C_sH_s)_n. The molar mass of this polystyrene can be calculated as: M(Br3C6H2(C_sH_s)_n) = M(Br)*3 + M(C)*6 + M(H)*2 + n * M(C_sH_s)
02

Calculate the mass percentage of bromine

The mass percentage of bromine can be calculated using the formula: Mass % of bromine = (mass of bromine in the polystyrene / total mass of polystyrene) * 100 The given mass percentage of bromine in the polystyrene is 10.46%. We can rewrite the formula as: 10.46 = (3*M(Br) / (M(Br)*3 + M(C)*6 + M(H)*2 + n * M(C_sH_s))) * 100 Given, M(Br)=80. Plug in the values in the formula: 10.46=((3*80)/(3*80+6*12+2*1+n*104))*100
03

Find the value of n

Now we need to solve the equation to find the value of n: 10.46=((240)/(240+72+2+n*104))*100 Divide both sides by 100: 0.1046= (240)/(314+n*104) Rearrange the equation to obtain n: n= (240/0.1046 - 314)/104 n ≈ 21 So, the correct answer is (b) 21.

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Most popular questions from this chapter

A quantity of \(0.25 \mathrm{~g}\) of a substance when vaporized displaced \(50 \mathrm{~cm}^{3}\) of air at \(0^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm} .\) The gram molecular mass of the substance will be (a) \(50 \mathrm{~g}\) (b) \(100 \mathrm{~g}\) (c) \(112 \mathrm{~g}\) (d) \(127.5 \mathrm{~g}\)

Iodobenzene is prepared from aniline \(\left(\mathrm{C}_{6} \mathrm{H}_{3} \mathrm{NH}_{2}\right)\) in a two-step process as shown here: \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}+\mathrm{HNO}_{2}+\mathrm{HCl} \longrightarrow\) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~N}_{2}^{+} \mathrm{Cl}^{-}+2 \mathrm{H}_{2} \mathrm{O}\) \(\mathrm{C}_{6} \mathrm{H}_{3} \mathrm{~N}_{2}^{+} \mathrm{Cl}^{-}+\mathrm{KI} \rightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{I}+\mathrm{N}_{2}+\mathrm{KCl}\) In an actual preparation, \(9.30 \mathrm{~g}\) of aniline was converted to \(16.32 \mathrm{~g}\) of iodobenzene. The percentage yield of iodobenzene is \((I=127)\)

An amount of \(0.3\) mole of \(\mathrm{SrCl}_{2}\) is mixed with \(0.2\) mole of \(\mathrm{K}_{3} \mathrm{PO}_{4}\). The maximum moles of \(\mathrm{KCl}\) which may form is (a) \(0.6\) (b) \(0.5\) (c) \(0.3\) (d) \(0.1\)

The volume strength of a sample of \(\mathrm{H}_{2} \mathrm{O}_{2}\) is \(8.96\) vol'. The mass of \(\mathrm{H}_{2} \mathrm{O}_{2}\) present in \(250 \mathrm{ml}\) of this solution is (a) \(0.4 \mathrm{~g}\) (b) \(27.2 \mathrm{~g}\) (c) \(6.8 \mathrm{~g}\) (d) \(108.8 \mathrm{~g}\)

The mass composition of universe may be given as \(90 \% \mathrm{H}_{2}\) and \(10 \% \mathrm{He}\). The average molecular mass of universe should be (a) \(2.20\) (b) \(2.10\) (c) \(3.80\) (d) \(3.64\)
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