115 
Calculated. 
I.  (C,iH,2N202)2H,SO,-6H20.        II.  (C2iH22NA)2H2SO/7H,0. 
H2O  12-351  per  cent.  14-062  per  cent. 
Found. 
I.  II. 
12-099  per  cent.    12-503  per  cent. 
From  this  it  is  seen  that  neutral  sulphate  of  strychnine  of  commerce, 
contains  six  molecules  of  water  of  crystallization. 
To  determine  the  solubility,  two  methods  were  employed,  namely  : 
that  of  direct  evaporation  of  a  weiglied  portion  of  a  saturated  solution  ; 
and  that  of  the  estimation  of  the  sulphuric  acid  in  the  salt,  as  baric 
sulphate.  The  first  determination  was  made  by  preparing  saturated 
solutions  of  each  salt,  at  a  temperature  of  15°C.,  taking  weighed  por- 
tions of  each  solution,  evaporating  to  dryness,  and  subjecting  the 
Tesidue  to  a  temperature  of  185°C.,  until  loss  in  weight  no  longer 
•occurred.  I.  11*078  gm.  of  solution  gave  0*2306  gm.  of  anhydrous 
sulphate  of  strychnine,  corresponding  to  0*2591  gm.  of  crystallized 
sulphate.  The  weight  of  the  salt  was  then  subtracted  from  the  weight 
of  solution,  and  gave  the  amount  of  water  required,  which  was  10*8189 
^m.,  and  the  following  equation  then  gives  the  number  of  parts  of 
water,  in  which  the  salt  is  soluble : 
0-2-590:10*8189::  l:x  or  41-76  parts  of  water. 
II.  12*21  gm.  of  saturated  solution  of  No.  2,  gave  0*2790  gm.  of 
<3rystillized  sulphate  of  strychnine,  which  after  further  calculation 
^ives  42*76  parts  of  water,  in  which  the  salt  is  soluble. 
I  next  proceeded  to  determine  tlie  solubility  by  the  barium 
method.  I  first  tested  the  saturated  solutions  with  litmus  paper  to  test 
their  neutrality,  and  found  both  to  be  exactly  neutral.  A  weighed 
portion  of  each  solution  saturated  at  15°C.  was  then  precipitated  by 
baric  chloride,  the  precipitate  washed,  dried,  ignited,  and  weighed. 
From  the  weight  of  baric  sulphate  so  obtained,  the  amount  of  sulphate 
of  strychnine  present  was  then  calculated,  and  from  this  the  amount  of 
water  required  to  dissolve  the  salt.  I.  9*3013  gm.  of  saturated  solution 
of  No.  1,  gave  0*0579  gm.  of  baric  sulphate,  and  the  following  equa- 
tion represents  the  corresponding  amount  of  strychnine  sulphate : 
BaSO,:  (C2iH,2N202)2H2B04  +  6H2O::  0*0579:  x  or 
233  :  874  ::  0-0579:  0.2177  gm. 
The  weight  of  sulphate  of  strychnine  0*2177  gm.  was  subtracted  from 
the  weight  of  solution  used,  which  gives  the  weight  of  Avater  used 
=9*0836  gm.  Then,  0*2177  :  9*0836 :  :  1  :  x  or  41*72  parts  of  water. 
II.  10*5815  gm.  of  saturated  solution  of  No.  2,  gave  0*0646  gm.  ot 
