Transcript I. The Nature of Solutions

Solutions
I
II
III
Concentration
Concentration Units
 The
amount of solute in a solution.
 Describing
Concentration
• % by mass - medicated creams
• % by volume - rubbing alcohol
• ppm, ppb - water contaminants
• molarity - used by chemists
• molality - used by chemists
Molarity
Molarity (M) =
M 
mol
L
moles solute
liters of solution
0.25
mol
0.25M

1L
PROBLEM: Dissolve 5.00 g of NiCl2•6 H2O (237.7 g/mol) in
enough water to make 250 mL of solution. Calculate the
Molarity.
Step 1: Calculate moles of
NiCl2•6H2O
5.00 g •
1 mol
237.7 g
= 0.0210 mol
Step 2: Calculate Molarity
0.0210 mol
0.250 L
= 0.0841 M
[NiCl2•6 H2O = 0.0841 M
What mass of oxalic acid, H2C2O4, is required
to make 250. mL of a 0.0500M solution?
Step 1: Change mL to L:250 mL * 1L/1000mL
= 0.250 L
Step 2: Calculate Moles
= (0.0500 mol/L) (0.250 L) = 0.0125 moles
Step 3: Convert moles to grams.
(0.0125 mol)(90.00 g/mol) = 1.13 g
Learning Check
How many grams of NaOH are required
to prepare 400. mL of 3.0 M NaOH
solution?
1) 12 g
2) 48 g
3) 300 g
Two Other Concentration Units
MOLALITY, m
m of solution =
mol solute
kilograms solvent
% by mass
grams
solute
% by mass =
grams solution
Percent Composition
 This
is the mass of the solute divided by
the mass of the solution (mass of solute
plus mass of solvent), multiplied by 100.
 Example: Determine the percent
composition by mass of a 100 g salt
solution which contains 20 g salt.
(20 g NaCl / 100 g solution) x 100 = 20% NaCl
solution
Molality
moles
of
sol
molality
(m)

kg
of
solv
0.25
mol
0.25m

1kg
mass of solvent only
1 kg water = 1 L water
Calculating Concentrations
Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H2O.
Calculate m & % of ethylene glycol (by mass).
Calculate molality
conc (molality)
=
1.00 mol glycol
0.250 kg H 2 O
 4.00 molal
Calculate weight %
% glycol
=
62.1 g
62.1 g
+ 250. g
x 100%
= 19.9%
Molality
 Find
the molality of a solution containing
75 g of MgCl2 in 250 mL of water.
75 g MgCl2
1 mol MgCl2
95.21 g MgCl2 0.25 kg water
m
mol
kg
= 3.2m MgCl2
Molality
 How
many grams of NaCl are req’d to make
a 1.54m solution using 0.500 kg of water?
0.500 kg water 1.54 mol NaCl
1 kg water
1.5
mol
1.5m

1kg
58.44 g NaCl
1 mol NaCl
= 45.0 g NaCl
Dilutions
 Preparation
of a desired solution by
adding water to a concentrate.
 Moles of solute remain the same.
M 1V1 = M 2V 2
Dilution
 What
volume of 15.8M HNO3 is required
to make 250 mL of a 6.0M solution?
GIVEN:
M1 = 15.8M
V1 = ?
M2 = 6.0M
V2 = 250 mL
= 0.250 L
WORK:
M1 V1 = M2 V2
(15.8M) V1 = (6.0M)(0.250 L)
V1 = 0.095 L of 15.8M HNO3