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Some Basic Concepts of Chemistry question

2022 · 25 Jun · Shift 1 · Q23
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Some Basic Concepts of Chemistry question

2022 · 25 Jun · Shift 1 · Q23

JEE MainChemistrySome Basic Concepts of ChemistryNumerical+4 / −1
Number of grams of bromine that will completely react with 5.0 g of pent-1-ene is ‾×\underline{\hspace{2cm}}\times​× 10 −-− 2 g. (Atomic mass of Br = 80 g/mol) [Nearest Integer]
Numerical answer
View written solutionFree

Correct answer: 1143

  1. Write the reaction

Pent-1-ene is an alkene with formula: C5H10\mathrm{C_5H_{10}}C5​H10​

Alkenes add bromine across the double bond in a 1:11:11:1 molar ratio: C5H10+Br2→C5H10Br2\mathrm{C_5H_{10} + Br_2 \rightarrow C_5H_{10}Br_2}C5​H10​+Br2​→C5​H10​Br2​

So, 111 mole of pent-1-ene reacts with 111 mole of Br2\mathrm{Br_2}Br2​.

  1. Find molar mass of pent-1-ene

M(C5H10)=5×12+10×1=60+10=70 g/molM(\mathrm{C_5H_{10}})=5\times 12 + 10\times 1 = 60+10=70\ \text{g/mol}M(C5​H10​)=5×12+10×1=60+10=70 g/mol

  1. Moles of pent-1-ene in 5.0 g

n(C5H10)=5.070=114 moln(\mathrm{C_5H_{10}})=\frac{5.0}{70}=\frac{1}{14}\ \text{mol}n(C5​H10​)=705.0​=141​ mol

  1. Moles of bromine required

Since the ratio is 1:11:11:1, n(Br2)=114 moln(\mathrm{Br_2})=\frac{1}{14}\ \text{mol}n(Br2​)=141​ mol

  1. Molar mass of bromine

Given atomic mass of Br = 808080 g/mol, so M(Br2)=2×80=160 g/molM(\mathrm{Br_2})=2\times 80=160\ \text{g/mol}M(Br2​)=2×80=160 g/mol

  1. Mass of bromine required

m(Br2)=n×M=114×160=11.42857 gm(\mathrm{Br_2})=n\times M = \frac{1}{14}\times 160 = 11.42857\ \text{g}m(Br2​)=n×M=141​×160=11.42857 g

  1. Match with the required format

The question says mass is: ‾×10−2 g\underline{\hspace{2cm}}\times 10^{-2}\ \text{g}​×10−2 g

So, 11.42857 g=1142.857×10−2 g11.42857\ \text{g} = 1142.857\times 10^{-2}\ \text{g}11.42857 g=1142.857×10−2 g

Nearest integer: 114311431143

Thus, the required integer is: 1143\boxed{1143}1143​

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