IIT JEE , AIEEE Forum - Chemistry , Physical Chemistry

s.mirchi

0.24 g of volatile substance upon vapourization gives 45ml vapour at NTP. what will be the vapour density of the substance.

tango_goat

use the equation PV=nRT
P=1atm
V=.045 L
R=.0821
T=273 K
n=.002.......thus .24g=.002 moles
mol wt=119.53
vapour density=1/2*mol wt
v.d=59.76 gm/l
is it correct?

somyekathait

let mol. weight be x

then at NTP 0.24/x = 45/22400 =moles

we find x =119.46

we know that Mol wt = 2X VD

hence VD = Mol Wt/2 =x/2 = 119.46/2= 59.73 (Answer)

rohitkuruvila

vapour density of a substance corresponds to 11200 ml,(molar volume/2)

0.24 g of substance gives 45 ml

x g gives 11200 ml

x=(11200*0.24)/45

x=59.73 g

the vapour density of the substance is 59.73

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