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Some Basic Concepts of Chemistry question

2022 · 26 Jul · Shift 2 · Q19
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Some Basic Concepts of Chemistry question

2022 · 26 Jul · Shift 2 · Q19

JEE MainChemistrySome Basic Concepts of ChemistryNumerical+4 / −1
A 100 mL100 \mathrm{~mL}100 mL solution of CH3CH2MgBr\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{MgBr}CH3​CH2​MgBr on treatment with methanol produces 2.24 mL2.24 \mathrm{~mL}2.24 mL of a gas at STP. The weight of gas produced is ‾\underline{\hspace{2cm}}​ mg. [nearest integer]
Numerical answer
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Correct answer: 3

  1. Identify the reaction

    Ethyl magnesium bromide reacts with methanol as a Grignard reagent:

    C2H5MgBr+CH3OH→C2H6+MgBrOCH3\mathrm{C_2H_5MgBr + CH_3OH \rightarrow C_2H_6 + MgBrOCH_3}C2​H5​MgBr+CH3​OH→C2​H6​+MgBrOCH3​

    So, the gas produced is ethane (C2H6)\left(\mathrm{C_2H_6}\right)(C2​H6​).

  2. Use the given gas volume at STP

    Given volume of gas:

    V=2.24 mL=2.24×10−3 LV = 2.24\ \text{mL} = 2.24 \times 10^{-3}\ \text{L}V=2.24 mL=2.24×10−3 L

    At STP, molar volume is:

    22.4 L mol−122.4\ \text{L mol}^{-1}22.4 L mol−1

    Therefore, moles of gas are:

    n=2.24×10−322.4=1.0×10−4 moln = \frac{2.24 \times 10^{-3}}{22.4} = 1.0 \times 10^{-4}\ \text{mol}n=22.42.24×10−3​=1.0×10−4 mol

  3. Calculate mass of ethane formed

    Molar mass of ethane:

    M(C2H6)=2(12)+6(1)=30 g mol−1M(\mathrm{C_2H_6}) = 2(12) + 6(1) = 30\ \text{g mol}^{-1}M(C2​H6​)=2(12)+6(1)=30 g mol−1

    Mass of gas:

    m=nM=(1.0×10−4)(30)=3.0×10−3 gm = nM = (1.0 \times 10^{-4})(30) = 3.0 \times 10^{-3}\ \text{g}m=nM=(1.0×10−4)(30)=3.0×10−3 g

  4. Convert into mg

    3.0×10−3 g=3.0 mg3.0 \times 10^{-3}\ \text{g} = 3.0\ \text{mg}3.0×10−3 g=3.0 mg

  5. Nearest integer

    3\boxed{3}3​

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