Preparation of Solutions
191
It is simple to prepare solutions of known molality. The volume of the final
solution doesn't enter into it at all.
PROBLEM:
Prepare a 2.00 m naphthalene (C 10 H 8 ) solution using 50.0 g CCI 4 as the solvent.
SOLUTION:
You are given the molality of the solution and the weight of the solvent, from
which you can find x , the number of moles of C 10 H 8 needed.
. . . ...
.v moles CipH g
Molallty =
m = o.osoo kg ecu =
2 '
00
x = 0.100 moles C 10 H 8 needed
Weight of C 10 H 8 needed = (0.100 moles C 10 H 8 )( 128
S ,
C '°
H> L )
\
mole C 10 H 8 /
= 12.8 g C 10 H 8
To prepare the solution, dissolve 12.8 g C 10 H 8 in 50.0 g CC1 4 . If you knew that the
density of CC1 4 is 1.59 g/ml, you could measure out
=3..4mlCCl 4
1.59 -ml
Mole Fraction
Another way of expressing concentrations that is used commonly with gases
(see p 162) and colligative properties (see p 328) is mole fraction, which is
defined (for a given component) as being the moles of component in question
divided by the total moles of all components in solution. For a solution that has
three components (A, B, and C), the mole fraction of A is given by
i c
r A ^
moles of A
mole fraction of A = X A = moles of A + moles of B + moles of C
If you think about it, it's also easy to make a solution of a given mole fraction.
PROBLEM:
Prepare a 0.0348 mole fraction solution of sucrose (C 12 H 2 2Oii, mole weight = 342
g/mole), using 100 g (that is, 100 ml) of water.
SOLUTION:
You are given the mole fraction of sucrose and the moles of water