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(C) Initial millimoles of acetic acid $= 100 \ mL 	imes 0.5 \ M = 50 \ mmol$.Millimoles of unadsorbed acetic acid neutralized by $NaOH = 40 \ mL 	im

Vedclass · Accepted Answer

$2500$

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(C) Initial millimoles of acetic acid $= 100 \ mL 	imes 0.5 \ M = 50 \ mmol$.Millimoles of unadsorbed acetic acid neutralized by $NaOH = 40 \ mL 	imes 1 \ M = 40 \ mmol$.Millimoles of acetic acid adsorbed on $1 \ g$ of charcoal $= 50 - 40 = 10 \ mmol = 10 	imes 10^{-3} \ mol$.Number of molecules of acetic acid adsorbed $= 10 	imes 10^{-3} \ mol 	imes 6.0 	imes 10^{23} \ mol^{-1} = 6.0 	imes 10^{21} \ 	ext{molecules}$.Surface area occupied by one molecule $= \frac{	ext{Total surface area}}{	ext{Number of molecules}} = \frac{1.5 	imes 10^2 \ m^2}{6.0 	imes 10^{21}} = 0.25 	imes 10^{-19} \ m^2 = 2500 	imes 10^{-23} \ m^2$.Comparing this with $P 	imes 10^{-23} \ m^2$,we get $P = 2500$.

Gas	Critical Temperature
$A$	$304 \ K$
$B$	$630 \ K$
$C$	$200 \ K$
$D$	$540 \ K$

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Gas Critical Temperature

$A$ $304 \ K$

$B$ $630 \ K$

$C$ $200 \ K$

$D$ $540 \ K$

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