Theory of Molecular Volumes. 

 1:2:4 Trichlorbenzene. C 6 H 3 C1 3 . 



X (Unsymmetrical or Odd). M.V. 149' 1. 



'' ]C1 B.P. 213°'0. 



989 



ox 



Cl, = aH s Cl a -aH,= 149-l-86-4 = 62-7. 



6^3 



1:2:4:5 TetracJilorhenzene. C fi TLCL. 



6^2' 



9J (Symmetrical or Even). M.V. 165'0. 



Ad B.P. 244° 0. 



C1 V" 

 ci 



C1 4 = 6 H 8 C1 4 -C 6 H 2 = 165-0 -83-2 = 81'8. 



1:2:3:4:5 Pentachlorbenzene. C 6 HC1 5 . 



9J (Unsymmetrical or Odd). M.V. 183*8. 



CI B.P. 276°-0. 



Cl 6 = C 6 HCl 6 -C 6 H = 183-8-80-0 = 103*8. 



Hea>acldorbenzene. CfiPL. 



CI 



Cl/\C1 (Symmetrical or Even). M.V. 200'0. 



B.P. 326°'0. 

 ■ClWCl - 



CI Cl 6 =200-0-76-8 = 123'2. 



If we subtract the values of the CI atoms we obtain the 

 following results : — 







Compound. 



n. 



V n CI. 



A. 



C G H.C1 



1 



2 



3 



4 

 5 

 6 



21-8 

 41-3 

 62-7 

 81-8 

 1038 

 1232 



19-5 



21-4 

 191 



22-0 

 194 



C 6 H 4 C1 2 



C G H 3 C1 3 



C f H Cl, 



C HC1 



C CI 





