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Some Basic Concepts of Chemistry question

2022 · 25 Jul · Shift 2 · Q20
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Some Basic Concepts of Chemistry question

2022 · 25 Jul · Shift 2 · Q20

JEE MainChemistrySome Basic Concepts of ChemistryNumerical+4 / −1
A sample of 4.5 mg of an unknown monohydric alcohol, R−OHR-OHR−OH was added to methylmagnesium iodide. A gas is evolved and is collected and its volume measured to be 3.1 mL. The molecular weight of the unknown alcohol is ‾\underline{\hspace{2cm}}​ g/mol. [Nearest integer]
Numerical answer
View written solutionFree

Correct answer: 33

  1. Reaction involved

A monohydric alcohol reacts with methylmagnesium iodide as:

ROH+CH3MgI→ROMgI+CH4ROH + CH_3MgI \rightarrow ROMgI + CH_4ROH+CH3​MgI→ROMgI+CH4​

So, 1 mole of alcohol gives 1 mole of methane gas.

  1. Given data
  • Mass of alcohol =4.5 mg=4.5×10−3 g= 4.5\text{ mg} = 4.5 \times 10^{-3}\text{ g}=4.5 mg=4.5×10−3 g
  • Volume of gas evolved =3.1 mL=3.1×10−3 L= 3.1\text{ mL} = 3.1 \times 10^{-3}\text{ L}=3.1 mL=3.1×10−3 L

Assuming gas volume is measured at STP, molar volume of a gas is:

22.4 L mol−122.4\text{ L mol}^{-1}22.4 L mol−1
  1. Calculate moles of methane evolved
n(CH4)=3.1×10−322.4n(CH_4)=\frac{3.1\times 10^{-3}}{22.4}n(CH4​)=22.43.1×10−3​ n(CH4)=1.384×10−4 moln(CH_4)=1.384\times 10^{-4}\text{ mol}n(CH4​)=1.384×10−4 mol

Since the stoichiometric ratio is 1:11:11:1,

n(ROH)=1.384×10−4 moln(ROH)=1.384\times 10^{-4}\text{ mol}n(ROH)=1.384×10−4 mol
  1. Calculate molar mass of the alcohol
M=massmoles=4.5×10−31.384×10−4M=\frac{\text{mass}}{\text{moles}}=\frac{4.5\times 10^{-3}}{1.384\times 10^{-4}}M=molesmass​=1.384×10−44.5×10−3​ M≈32.5 g/molM \approx 32.5\text{ g/mol}M≈32.5 g/mol

Nearest integer:

33\boxed{33}33​
  1. Comparison with stored answer

Stored correct answer = 333333

My derived answer = 333333

So they agree.

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