CHEMISTRY 264 :
ANALYTICAL CHEMISTRY
TUTORIAL I :
the mol substance litre
molarity amount of per solution
:
c
=
mol/dm3 or c =
mal/ L
the of substance solvent
molality amount of per
kg
:
b =
mol/kg
molarity depends on volume of the solution
,
which
changes
with temperature due to thermal expansion
molality depends
the the solvent which does with
on mass of ,
not
change
temperature
(i) water
0 1 0013
glmL
=
,
for L :
p
A
=
M =
1 , 0013 x 1000 mL
1001
=
,
39
molar mass=18 , 015 glma
n =
Is sa
=
=
=
55 , 58
=
55 , 58 M
,(ii) benzene
0 , 8765
0 g(mL
=
for 1 L :
p
A
=
m =
0 , 8765x1000 mL
=
876 ,
59
molar mass =
78 , 11 glmal
n =
= cu
=
11 , 22
=
11 , 22 M
(ii) acetonitrile
0 , 783
0 g/mL
=
for 1 L :
p
A
=
M =
0 , 783 x 1000 mL
783
=
9
molar mass =
78 , 11 glmal
n =
Ec
=
19 08 M
19 08
=
= ,
,
, HBr : 31 % We
wt =
A solute x 10
of the solution is
assuming the total mass log
solute
31 =
sof x0
mass of solute
319
=
density 1 22 kg
:
,
p =A
mass mass of solution 1009
319
=
=
0
0
031k9 1kq
= =
,
,
m
i
n =
it
=
+ 79 , 904
=
0 , 38 Mol
=
, 08
0 L
molarity :
C =
os
=
4 75
, mol/L
mass of solvent :
molality : b
=3q
1009 319
=
-
699 5 51
=
=
,
ma/kg
069kg
= ,
0
, ppb = 800
ug/
assume a volume of 1 litre
M(CaHio0) 150 , 177
g/mol
=
-
n
=
=
0 , 000 00532704
molarity
:
c =
x000005327
=
=
5, 33x10-3
ANALYTICAL CHEMISTRY
TUTORIAL I :
the mol substance litre
molarity amount of per solution
:
c
=
mol/dm3 or c =
mal/ L
the of substance solvent
molality amount of per
kg
:
b =
mol/kg
molarity depends on volume of the solution
,
which
changes
with temperature due to thermal expansion
molality depends
the the solvent which does with
on mass of ,
not
change
temperature
(i) water
0 1 0013
glmL
=
,
for L :
p
A
=
M =
1 , 0013 x 1000 mL
1001
=
,
39
molar mass=18 , 015 glma
n =
Is sa
=
=
=
55 , 58
=
55 , 58 M
,(ii) benzene
0 , 8765
0 g(mL
=
for 1 L :
p
A
=
m =
0 , 8765x1000 mL
=
876 ,
59
molar mass =
78 , 11 glmal
n =
= cu
=
11 , 22
=
11 , 22 M
(ii) acetonitrile
0 , 783
0 g/mL
=
for 1 L :
p
A
=
M =
0 , 783 x 1000 mL
783
=
9
molar mass =
78 , 11 glmal
n =
Ec
=
19 08 M
19 08
=
= ,
,
, HBr : 31 % We
wt =
A solute x 10
of the solution is
assuming the total mass log
solute
31 =
sof x0
mass of solute
319
=
density 1 22 kg
:
,
p =A
mass mass of solution 1009
319
=
=
0
0
031k9 1kq
= =
,
,
m
i
n =
it
=
+ 79 , 904
=
0 , 38 Mol
=
, 08
0 L
molarity :
C =
os
=
4 75
, mol/L
mass of solvent :
molality : b
=3q
1009 319
=
-
699 5 51
=
=
,
ma/kg
069kg
= ,
0
, ppb = 800
ug/
assume a volume of 1 litre
M(CaHio0) 150 , 177
g/mol
=
-
n
=
=
0 , 000 00532704
molarity
:
c =
x000005327
=
=
5, 33x10-3
