Question

What are the normality and molarity of a solution prepared by dissolving 8.050g Ba(OH)2•8H2O in 1500ml of solution?
M = mols/L = 8.050/molar mass Ba(OH)2 /1.500 L = ?
Technically, one cannot determine the equivalent weight without know the reaction/titration; however, usually when no reaction is given we count the number of H or OH in the molecule. In the case of Ba(OH)2 it would be eq. wt. = molecular weight/2, then
N = # equivalents/L so that would be
N = (8.050 g/molar mass/2)/1.500

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