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
