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Some Basic Concepts of Chemistry question

2025 · 3 Apr · Shift 2 · Q21
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Some Basic Concepts of Chemistry question

2025 · 3 Apr · Shift 2 · Q21

JEE MainChemistrySome Basic Concepts of ChemistryNumerical+4 / −1
X g of nitrobenzene on nitration gave 4.2 g of m -dinitrobenzene. X = ‾\underline{\hspace{2cm}}​ g. (nearest integer) [Given : molar mass (in gmol−1)C:12,H:1,O:16, N:14]\left.\left.\mathrm{g} \mathrm{mol}^{-1}\right) \mathrm{C}: 12, \mathrm{H}: 1, \mathrm{O}: 16, \mathrm{~N}: 14\right]gmol−1)C:12,H:1,O:16, N:14]
Numerical answer
View written solutionFree

Correct answer: 3

  1. Write the reaction idea

Nitrobenzene on further nitration gives m-dinitrobenzene:

Nitrobenzene→m-dinitrobenzene\text{Nitrobenzene} \rightarrow \text{m-dinitrobenzene}Nitrobenzene→m-dinitrobenzene

This is a 1:11:11:1 molar conversion:

1 mol nitrobenzene→1 mol m-dinitrobenzene1\ \text{mol nitrobenzene} \to 1\ \text{mol m-dinitrobenzene}1 mol nitrobenzene→1 mol m-dinitrobenzene

  1. Find molar mass of nitrobenzene

Nitrobenzene has formula C6H5NO2\mathrm{C_6H_5NO_2}C6​H5​NO2​.

M(C6H5NO2)=6(12)+5(1)+14+2(16)M(\mathrm{C_6H_5NO_2}) = 6(12) + 5(1) + 14 + 2(16)M(C6​H5​NO2​)=6(12)+5(1)+14+2(16)

=72+5+14+32=123 g mol−1= 72 + 5 + 14 + 32 = 123\ \mathrm{g\ mol^{-1}}=72+5+14+32=123 g mol−1

  1. Find molar mass of m-dinitrobenzene

m-Dinitrobenzene has formula C6H4N2O4\mathrm{C_6H_4N_2O_4}C6​H4​N2​O4​.

M(C6H4N2O4)=6(12)+4(1)+2(14)+4(16)M(\mathrm{C_6H_4N_2O_4}) = 6(12) + 4(1) + 2(14) + 4(16)M(C6​H4​N2​O4​)=6(12)+4(1)+2(14)+4(16)

=72+4+28+64=168 g mol−1= 72 + 4 + 28 + 64 = 168\ \mathrm{g\ mol^{-1}}=72+4+28+64=168 g mol−1

  1. Calculate moles of m-dinitrobenzene formed

Given mass of m-dinitrobenzene =4.2 g= 4.2\ \mathrm{g}=4.2 g.

n=4.2168=0.025 moln = \frac{4.2}{168} = 0.025\ \mathrm{mol}n=1684.2​=0.025 mol

  1. Use 1:1 mole ratio

So moles of nitrobenzene used =0.025 mol= 0.025\ \mathrm{mol}=0.025 mol.

  1. Calculate mass of nitrobenzene

X=0.025×123=3.075 gX = 0.025 \times 123 = 3.075\ \mathrm{g}X=0.025×123=3.075 g

Nearest integer:

X≈3 gX \approx 3\ \mathrm{g}X≈3 g

Final Answer

3\boxed{3}3​

The derived answer matches the stored correct answer.

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