How many grams of Fe(NO3)3 · 9 H2O are required to prepare 200.0 mL of a 0.260 M solution of Fe(NO3)3 · 9 H2O?
1) How many grams of Fe(NO3)3 · 9 H2O are required to prepare 200.0 mL of a 0.260 M solution of Fe(NO3)3 · 9 H2O?
2) What is the molarity of the resulting solution when 25.674 g of Mn(ClO4)2 · 6 H2O are added to 200.0 mL of water?
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? g Fe(NO3)3*9H2O = 0.2000L x 0.260 mole/L x 404 g/mol Fe(NO3)3*9H2O = 21.0 g
mole Mn(ClO4)2*6H2O = 25.674 g /361.9 g Mn(ClO4)2*6H2O/mol = 0.07094 mol
molality = 0.07094 mole/0.200 kg H2O = 0.3547 m
Fe No3