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

Another expression with which one should be familiar is millimoles per liter

(mmol/L). It is a variation of the mole/liter concentration unit.

a. A millimole = 0.001 mole and 1 mole = 1000 millimoles.

MD0837

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