b. Example 2. What volume of CaCO3 solution can be prepared from 25.0 g of
CaCO3 if a 0.50 mol/L CaCO3 is desired?
Solution. Read the problem carefully and determine the desired quantity.
Volume of CaCO3 solution.
Calculate the GMW of the compound.
CaCO3
Ca
40.1 X 1 = 40.1
C
12.0 X 1 = 12.0
O
16.0 X 3 = + 48.0
100.1 g/mol
Determine the number of moles contained in 25.0 grams of solute.
1 mol
25.0 g X -------- = 0.250 mol
100.1 g
Use ratio and proportion to determine the amount of solution that can be
prepared.
0.50 mol
0.250 mol
-------- = ------------
1L
xL
(0.50 mol)(x L) = (0.250 mol)(1 L)
(0.250 mol)(1 L)
x L = ---------------- = 0.50 L
0.50 mol
Section III. MILLIMOLES PER LITER
3-6.
INTRODUCTION
Another expression with which one should be familiar is millimoles per liter
(mmol/L). It is a variation of the mole/liter concentration unit.
3-7.
DISCUSSION
a. A millimole = 0.001 mole and 1 mole = 1000 millimoles.
MD0837
3-6