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Practical Organic Chemistry question

2025 · 2 Apr · Shift 1 · Q22
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Practical Organic Chemistry question

2025 · 2 Apr · Shift 1 · Q22

JEE MainChemistryPractical Organic ChemistryNumerical+4 / −1
0.1 mol of the following given antiviral compound (P)(\mathrm{P})(P) will weigh ‾\underline{\hspace{2cm}}​×10−1 g\times 10^{-1} \mathrm{~g}×10−1 g(nearest integer). JEE Main 2025 (Online) 2nd April Morning Shift Chemistry - Practical Organic Chemistry Question 8 English (Given : molar mass in gmol−1H:1,C:12, N:14,O:16, F:19,I:127\mathrm{g} \mathrm{mol}^{-1} \mathrm{H}: 1, \mathrm{C}: 12, \mathrm{~N}: 14, \mathrm{O}: 16, \mathrm{~F}: 19, \mathrm{I}: 127gmol−1H:1,C:12, N:14,O:16, F:19,I:127 )
Numerical answer
View written solutionFree

Correct answer: 157

  1. Read the molecular structure

    The antiviral compound shown is Favipiravir, whose molecular formula is: C5H4FN3O2\mathrm{C_5H_4FN_3O_2}C5​H4​FN3​O2​

  2. Calculate its molar mass

    Using the given atomic masses:

    • Carbon: 5×12=605 \times 12 = 605×12=60
    • Hydrogen: 4×1=44 \times 1 = 44×1=4
    • Fluorine: 1×19=191 \times 19 = 191×19=19
    • Nitrogen: 3×14=423 \times 14 = 423×14=42
    • Oxygen: 2×16=322 \times 16 = 322×16=32

    Therefore, M=60+4+19+42+32=157 g mol−1M = 60 + 4 + 19 + 42 + 32 = 157\ \text{g mol}^{-1}M=60+4+19+42+32=157 g mol−1

  3. Mass of 0.10.10.1 mol

    m=nM=0.1×157=15.7 gm = nM = 0.1 \times 157 = 15.7\ \text{g}m=nM=0.1×157=15.7 g

  4. Match with the required form

    We need 15.7 g=157×10−1 g15.7\ \text{g} = 157 \times 10^{-1}\ \text{g}15.7 g=157×10−1 g

    Nearest integer =157= 157=157.

  5. Compare with stored answer

    My derived answer is 157, but the stored correct answer is 372.

    So, I do not agree with the stored answer.

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