Transcript Topic 1 Solution Concentration

 solute – the part of a
solution that is being
dissolved (usually the
lesser amount)
 solvent – the part of a
solution that dissolves
the solute (usually the
greater amount)
 solution = solute +
solvent
Solute Solvent
Solution
Si
Al
metal alloy
salt
water
salt water
CO2
water
carbonated
water
O
N
air
 a measure of the amount of solute that
is dissolved in a given quantity of
solvent
 two ways to measure
1. Molarity (M) which is moles/volume
 volume must be in liters
 IB uses dm3 for liter
 can use square brackets to mean
concentration
 [HCl] = 1.0 M
 The number of moles of solute dissolved in
liters of solution
Molarity =
moles solute
liters of solution
1.0 L of water was used to make 1.0 L of solution. Notice
the water left over. 1.0L of water would be too much.
top off
to 1 liter
 What is the concentration of 2 moles of
HCl in 4 dm3?
 2 mol / 4 dm3 = 0.5 M
Step 1: Calculate moles
of NiCl2• 6H2O in 5 g of
NiCl2• 6H2O
1 mol
5.00 g •
= 0.0210 mol
237.7 g
Step 2: Calculate Molarity
0.0210 mol
= 0.0841 M
0.250 L
1.0 L = 1000 mL
.25 L = 250 mL
What mass of oxalic acid, H2C2O4, is required to
make 250 mL of a 0.0500 M solution?
M = moles/V  moles = M•V
Step 1: Change mL to L.
250 mL/1000 = 0.250 L
Step 2: Calculate.
moles = (0.0500 M) (0.250 L) = 0.0125 moles H2C2O4
Step 3: Convert moles to grams.
(0.0125 mol H2C2O4) (90.00 g) =
1 mol H2C2O4
1.13 g
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
2. Molality which is mass/volume

grams/liters
 or IB says g dm-3
 What is the concentration of 15 g of NaCl
dissolved in 500 ml of water
 15g/.5L = 30 g dm-3