The mass (in grams) of FeSO4•7H2O (The dot means it is still a part of the whole compound, so include the water) required for preparation of 125 mL of 0.90 M solution is

Answer 1

The mass (in grams) of FeSO4•7H2O (The dot means it is still a part of the whole compound, so include the water) required for preparation of 125 mL of 0.90 M solution is

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You first need to find the Molecular Weight of FeSO4•7H2O

One mole of FeSO4*7H20 has the mass of the sum of all the individual atoms' atomic weights that can be found on the periodic table.

55.845 + 32.06 + 15.99*4 + 7*1*2 + 7 * 15.99 ~= 278.01 grams

That's is weight of 1 mole of FeSO4*7H20

concentrations are measured in Molarity or moles/Liters

Your solution needs to be 0.9M in concentration and 125 mL in volume.

so you need

0.9 mol/L* (278.01g/mol) = 250.209g/L == 250.209g/1000mL

But you only need 125 mL of it.

so 125 mL * (250.209g/1000mL) ~= 31.28 grams

is the amount of FeSO4 7H20 that needs to dissolved in 125 mL of water.