Chemistry - Mole Concept Question with Solution | TestHub

ChemistryMole ConceptAtomic mass/Average Atomic Mass/Average Molecular MassMedium2 minQB

ChemistryMediumsingle choice

Passage / Comprehension

According to Avogadro's law, equal numbers of moles of gases occupy the same volume under identical conditions of temperature and pressure. Even if we have a mixture of non-reacting gases, Avogadro's law is still obeyed by assuming the mixture as a new gas.

Now, let us assume air consists of 80% by volume of Nitrogen (N₂) and 20% by volume of Oxygen (O₂). If air is taken at STP, then 1 mole of air would occupy 22.4 L. One mole of air would contain 0.8 mol of N₂ and 0.2 mol of O₂, hence the mole fractions of N₂ and O₂ are given by $X_{N_{2}} = 0.8$ and $X_{O_{2}} = 0.2$ .

Density of air at STP is

Options:

Answer:

Solution:

$m_{ai r} = m_{N_{2}} + m_{O_{2}}$

$1 \times M_{ai r} = 0.8 \times 28 + 0.2 \times 32 = 28.8$

$a_{ai r} = \frac{P \cdot M _{ai r}}{RT} = \frac{28.8}{22.4} = \frac{9}{7}$

Stream:JEESubject:ChemistryTopic:Mole ConceptSubtopic:Atomic mass/Average Atomic Mass/Average Molecular Mass

⏱ 2mℹ️ Source: QB

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