Calculate the amount of salt needed using the appropriate factors.
----------X ----------X 0.100 L = 301 mg
4-10. PROBLEMS INVOLVING HYDRATES
Like molar solutions, normal solutions contain a known number of particles per
liter of solution. An one equivalent per liter (Eq/L) solution of CuSO4 will have the same
concentration as a one equivalent per liter CuSO4.5H2O solution. When solving normal
problems involving hydrates, use the gram equivalent weight of the substance being
weighed in the preparation of the solution.
How much CuSO4.5H2O is required to prepare 500 mL of a 1.50 Eq/L CuSO4 solution?
Solution. Read the problem carefully and determine the desired quantity.
Grams of CuSO4.5H2O.
Determine the GEW of the compound that is available (actually being weighed
63.5 X 1 = 63.5
32.1 X 1 = 32.1
16.0 X 4 = 64.0
18.0 X 5 = + 90.0
When determining the TPIV of hydrates, the water (H2O) molecules are not
considered. Water has a low degree of reactivity. Also, note that copper is
ambivalent, and we must consider the charge of the anion (-) in order to
correctly determine the TPIV.
-------- X ------ = 124.8 g/Eq