Summary Analytical Chemistry (Chemistry 264) | Worked Tutorials, Quizzes & Lecture Slide Questions

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