Chemistry - Mole Concept Question with Solution | TestHub

ChemistryMole ConceptCalculation of moleMedium2 minQB

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The treatment of $CH_{3} OH$ with $CH_{3} Mgl$ releases 1.04 ml of a gas at STP . The mass of $CH_{3} OH$ added is

Answer:

Solution:

CH₃OH + CH₃MgI $⟶$ CH₄ + CH₃OMgI

Moles of CH₄ = $\frac{1.4}{22400} = 4.642 \times 1 0^{- 5}$

Number of moles of CH₃OH = $4.642 \times 1 0^{- 5}$ .

Mass of CH₃OH = $4.642 \times 1 0^{- 5} \times 32 = 1.485$ mg.

Stream:JEESubject:ChemistryTopic:Mole ConceptSubtopic:Calculation of mole

⏱ 2mℹ️ Source: QB

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