Describe the preparation of 400 mL of a solution that is 0.0810M in K+, starting with solid K4Fe(CN)6

There are 4 moles of K+ in each mole of K4Fe(CN)6. Therefore a solution of x M of K4Fe(CN)6 would be 4*x M in K+. The molarity of K4Fe(CN)6 should be 1/4 the desired molarity of K+, So you need to prepare a solution that is 0.0810/4 = 0.02025 M in K4Fe(CN)6. Determine the molecular mass of K4Fe(CN)6:

The molecular mass of the elements involved are

K = 39.10
Fe = 55.85
C = 12.01
N = 14.01

The total for K4Fe(CN)6 is then 4*39.10 + 55.85 + 6*12.01 * 6*14.01 = 368.4
A 0.02025 M solution would then be 0.02025*368.4 g of K4Fe(CN)6 in one L of solvent. Since we want 400 mL or 0.4L, the amount of solute requred is

0.400*0.0.02025*368.4 = 2.968 g

Add solvent to 2.968 g of K4Fe(CN)6 to make 400 mL of solution; you will have 0.02025 M solution of the compound, and 4x this or 0.0810 M of K+