Chemistry - Mole Concept Question with Solution | TestHub

ChemistryMole ConceptExpetimental Determination of Atomic Mass/Molecuair massEasy2 minQB

ChemistryEasymultiple choice

The density of air is $0.001293 g / cm^{3} at 1 atm$ and 273 K . Identify which of the following statement is correct?

Answer:

A, B

Solution:

Convert density to g/L: Density of air at STP = $0.001293 \times 1000 = 1.293 g / L$ .

Calculate Molecular Weight: $M = De n s i t y \times M o l a r V o l u m e = 1.293 \times 22.4 \approx 28.96 g / m o l$ (Statement B is correct).

Calculate Vapour Density: $Va p o u rDe n s i t y = \frac{M o l ec u l a r W e i g h t}{2} = \frac{28.96}{2} = 14.48$ (Statement A is correct).

Conclusion: Vapour density is a dimensionless unit, meaning options (A) and (B) are the correct statements.

Stream:JEESubject:ChemistryTopic:Mole ConceptSubtopic:Expetimental Determination of Atomic Mass/Molecuair mass

⏱ 2mℹ️ Source: QB

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