Multiply the desired volume in liters times the millimolar concentration times

the millimolar weight to determine the amount of salt needed to prepare the solution.

300 mmol

58.5 mg

0.250 L X ---------- X ---------- = 4390 mg

L

mmol

Since molarity is based on the number of moles (Avogadro's number) per liter of

solution, then one mole per liter of CuSO4 will contain the same number particles as one

mole per liter of CuSO4 5H2O. When calculating molarity problems involving hydrates,

use the gram molecular weight of the substance being used (weighed) in the

preparation of the solution.

Example**. **How much CuSO4 5H2O is required to prepare 250 mL of a 2.00 mol/L

CuSO4 solution?

Solution. Read the problem carefully and determine the desired quantity.

Grams of CuSO4 5H2O.

Determine the GMW of the available compound.

CuSO4 5H2O

Cu 63.5 X 1 = 63.5

S

32.1 X 1 = 32.1

O 6.0 X 4 =

64.0

H

1.0 X 10 = 10.0

O 16.0 X 5 = + 80.0

249.6 g/mol

Express the desired volume in liters.

1L

250 mL X ---------- = 0.250 L

1000 mL

MD0837

3-9

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