Vapour density of pure PCl 5 is 104.16 , but when heated to 23 0 ∘ C , its vapour density is reduced to 62 . The degree of dissociation of PCl 5 at 23 0 ∘ C is

α = D D − d = 62 104.16 − 62 = 0.68 O r 68%

Chemistry - Chemical Equilibrium Question with Solution | TestHub

ChemistryChemical EquilibriumDegree of Dissociation/Vapour Density/Molar MassMedium2 minQB

ChemistryMediumsingle choice

Vapour density of pure $PCl_{5}$ is 104.16 , but when heated to $23 0^{\circ} C$ , its vapour density is reduced to 62 . The degree of dissociation of $PCl_{5}$ at $23 0^{\circ} C$ is

Options:

Answer:

Solution:

$α = \frac{D - d}{D} = \frac{104.16 - 62}{62} = 0.68 O r 68%$

Stream:JEESubject:ChemistryTopic:Chemical EquilibriumSubtopic:Degree of Dissociation/Vapour Density/Molar Mass

⏱ 2mℹ️ Source: QB

Doubts & Discussion

Loading discussions...