Preparation of Solutions
193
SOLUTION:
(a) To find the molality we need to know, for a given amount of solution, the
moles of H 2 SO 4 and the kg of H 2 O. If we take a liter of solution, we shall have
1840 g of solution, of which 4.0% is water, so (0.040)( 1840 g) = 74 g H 2 O.
Because there are 18.0 moles of H 2 SO 4 in this liter, we have
(b) To find the mole fraction, we need to know (for a given amount of solution)
the moles of H 2 SO 4 and the moles of H 2 O. If we take a liter of solution, we
shall have 18.0 moles of H 2 SO 4 and
74 g "
2
° = 4.1 moles of H 2 O
,o
mole H 2 O
Therefore the mole fraction is
y
=
_
l8.0molesH 2 S0 4
__
*
H '
S<)4
(18.0 moles H 2 SO 4 ) + (4.1 moles H 2 0)
Note that it is not possible to convert from molarity to molality or mole
fraction unless some information about the density or weight composition of
the solution is given.
Dilution
One of the most common ways to prepare a solution is to dilute a concentrated
solution that has already been prepared. There is a fundamental principle that
underlies all dilutions: the number of moles of solute is the same after dilution
as before. It is only the moles of solvent that have been changed (increased).
This principle makes dilution calculations simple. If M l and M 2 are the
molarities before and after dilution, and V l and V 2 are the initial and final
volumes of solution, then
moles of solute before dilution = moles of solute after dilution
moles\
/
,, ,.
liter)
( Vi llters) = (
M * iteT(V *
llters)
The following problem illustrates the use of this equation.
PROBLEM:
What volume of 18.0 M H 2 SO 4 is needed for the preparation of 2.00 liters of 3.00 vi
H 2 SO 4 ?