After making this more of a sample problem, I believe the answer is #1.14 (mol)/(dm^3)# (assuming 100 grams of water).

The solution's percent concentration by mass is defined as

#c% = (mass_(solute))/(mass_(t otal solution)) * 100#, where

#mass_(t otal solution) = mass_(solvent) + mass_(solute)#

Now, let us assume that we have #mass_(solvent) = 100g# of water, #H_2O#,

#c% = (mass_(solute))/(mass_(solute) + 100) * 100# = #21.80#.

We would then have

#mass_(solute) = 0.218 * (mass_(solute) + 100)#, which means

#mass_(solute) = 27.9grams#, the solute being of course #K_2CrO_4#.

The number of moles of #K_2CrO_4# can be determined by

#n_(K_2CrO_4) = (27.9 g) / (194.2 g/(mo l e)) = 0.14 mol es#, where #194.2 g/(mo l e)# is the molar mass of the solute.

Knowing the density of the solute, we can determine its volume

#V_(K_2CrO_4) = m/(density) = (27.9 g)/(1.20 g/(cm^3)) = 23.3 cm^3#

Knowing that #1 dm^3# = #10^(3) cm^3#, we get a volume of #V = 0.023 dm^3#.

The density of water is #1g/(cm^3)#, therefore the volume of water is equal to

#V_(H_2O) = (100g)/(1 g/(cm^3)) = 100 cm^3 = 0.1 dm^3#

Therefore, the molarity can be calculated by

molarity # = (n_(K_2CrO_4))/(V_(t otal)) = 0.14/(0.023+0.1) = 1.14 M#