190
Stoichlometry III: Calculations Based on Concentrations of Solutions
comes from the weighed sample is irrelevant. The source of the water is never a
matter of concern.
Many times it is convenient to obtain a substance from its solution. What
volume of solution should you use to get the quantity of solute you want?
PROBLEM:
What volume of 0.250 M Na 2 CrO 4 will be needed in order to obtain 8.10 g of
Na 2 CrO 4 ?
SOLUTION:
If V = liters of solution needed, it must supply the number of moles contained in
8.10 g of Na 2 CrO 4 .
Moles of Na 2 Cr0 4 needed =
8 -
10 gNa 2 CrO 4 = QQ5QQ ^^
1620 _gNa 2 CrO ± _
mole NajCrO.,
Moles of Na 2 Cr0 4 in V liters = (o.250 mo'^N^CrOA (V liters)
= 0.250V moles
0.250V = 0.0500
V = 0.200 liter = 200 ml needed
This problem illustrates the two most common ways of calculating moles of a
compound: (a) weight divided by mole weight, and (b) molarity times volume in
liters.
Molality
When discussing the colligative properties of a solution (Chapter 21), it is more
important to relate the moles of solute to a constant amount of solvent rather
than to the volume of the solution, as in the case of molarity. In practice this is
accomplished by using a kilogram of solvent instead of a liter of solution as the
reference. A solution that contains one mole of solute per kilogram of solvent is
known as a one molal solution; it is abbreviated 1.00 m. In general,
molality of solution = m=
of solute
kg or solvent
Because the density of water is approximately 1 g/ml, the molarities and
molalities of water solutions will have about the same value. This will not be
true for most other solvents.
Stoichlometry III: Calculations Based on Concentrations of Solutions
comes from the weighed sample is irrelevant. The source of the water is never a
matter of concern.
Many times it is convenient to obtain a substance from its solution. What
volume of solution should you use to get the quantity of solute you want?
PROBLEM:
What volume of 0.250 M Na 2 CrO 4 will be needed in order to obtain 8.10 g of
Na 2 CrO 4 ?
SOLUTION:
If V = liters of solution needed, it must supply the number of moles contained in
8.10 g of Na 2 CrO 4 .
Moles of Na 2 Cr0 4 needed =
8 -
10 gNa 2 CrO 4 = QQ5QQ ^^
1620 _gNa 2 CrO ± _
mole NajCrO.,
Moles of Na 2 Cr0 4 in V liters = (o.250 mo'^N^CrOA (V liters)
= 0.250V moles
0.250V = 0.0500
V = 0.200 liter = 200 ml needed
This problem illustrates the two most common ways of calculating moles of a
compound: (a) weight divided by mole weight, and (b) molarity times volume in
liters.
Molality
When discussing the colligative properties of a solution (Chapter 21), it is more
important to relate the moles of solute to a constant amount of solvent rather
than to the volume of the solution, as in the case of molarity. In practice this is
accomplished by using a kilogram of solvent instead of a liter of solution as the
reference. A solution that contains one mole of solute per kilogram of solvent is
known as a one molal solution; it is abbreviated 1.00 m. In general,
molality of solution = m=
of solute
kg or solvent
Because the density of water is approximately 1 g/ml, the molarities and
molalities of water solutions will have about the same value. This will not be
true for most other solvents.