Chemistry - Surface Chemistry Question with Solution | TestHub

ChemistrySurface ChemistryAdsortionMedium2 minQB

ChemistryMediuminteger

Graph between $lo g (\frac{X}{m})$ vs $lo g P$ is provided for adsorption of $NH_{3}$ gas on metal surface. Calculate weight of $NH_{3}$ (in gram) gas adsorbed by 20 gm of metal surface at 2 atm pressure

Answer:

120

Solution:

$lo g (\frac{x}{m}) = lo g K + \frac{1}{n} lo g P \Rightarrow \frac{1}{n} = 1$

$lo g K = 0.48 = lo g 3$

$K = 3$

$\frac{x}{m} = K p^{1/ n}$

$\frac{x}{20} = 3. (2)^{1}$

$x = 120 gm$

Stream:JEESubject:ChemistryTopic:Surface ChemistrySubtopic:Adsortion

⏱ 2mℹ️ Source: QB

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