DOC.
639
OCTOBER
1918
677
then
4.4
g Na2SO4
exist in
95.6
g
water. Since the
specif. weight
of
the
10%
Glauber’s
salt
solution
is
1.040,
100
g
of
such
a
solution
thus
has
the
volume
100/1.040 =
96.15
cc.
Of
this, 95.6
cc
is
water,
so
for
the
4.4
g Na2SO4,
0.55
cc
remain. The
specif,
wt. of
anhydr. Na2SO4
would therefore be
4.4/0.55
=
8.-
The
specif,
weight
of
a
10%
solution
of
Na2SO4
determined from
this
quantity
is
1.095,
which
agrees
well
with
the
observed value
1.093.
Likewise,
the determination
of the
specif,
wt.
of
a
20%
solution of
Na2SO4
+
10
H2O,
interpolated
in
the
same way
from
the
10% solution,
yields 1.083;
the
observed
sp.
w.
is 1.082.
The
friction observations
performed
at
my
instigation by a
chemist at
the
military
base
hospital,
then
on
active
duty
as a
soldier,
refuse
to
agree
with this
at
all,
however. The
data
on
the
change
in volume for
the
solution
already
are
striking. Upon dissolving
10
g
of Glauber’s salt into
90
g
water
(at
15°),
the total
volume of
the
solution is
supposed
to
have amounted to
only
92.5
cc,
whereas
according
to
the
specific weight,
(1.040)
it would have
to
be
96.15
cc.-Since
the table
leaves
an
unreliable
impression
otherwise
as
well
and since
it
was
not
possible
for
me
to
come
up
with
any
kind of
system
behind
it,
I
shall refrain from
reproducing
it.-
The
behavior
of
the
similarly
structured
Epsom
salt
MgSO4
+
7H2O
is
very
interesting;
specif.
wt.
of
the
dry
substance
1.685,
molar wt.
261
(of
this
135
to
MgSO4
and
126
to
7 H2O).-At 13°5
n
for water
50.4
s.
2g
Epsom
salt
98
cc
Water:
vol
ca.
99
cc
n'
=
53.3
s.
Temp. 16°5
in
of
solution
4
” ”
96
”
"
"
ca.
98
” ”
56.6
s.
" "
6 94
ca. 971/4
59.1 " "
10 90
ca.
951/2
64
" "
14
86
ca.
931/2
69.5 "
"
16 84
ca.
923/4
75 " "
20 80
ca.
911/4
85 " "
If
the
difference in volume
is
entered into
your
formula
as
V',
the
observed values
for
n'
come
out
quite
well,
if
K
is
set
equal
to
5
instead of to
2.5.-[10]
The
factor