What was the resulting molarity of your primary standard solution of potassium iodate kio3?

Missing question: 0,535 gram of KIO₃ dissolved in 250 mL of de-ionized water to make primary standard solution.
m(KIO₃) = 0,535 g.
V(KIO₃) = 250 mL ÷ 1000 mL/L = 0,25 L.
n(KIO₃) = m(KIO₃) ÷ M(KIO₃).
n(KIO₃) = 0,535 g ÷ 214 g/mol.
n(KIO₃) = 0,0025 mol.
c(KIO₃) = n(KIO₃) ÷ V(KIO₃).
c(KIO₃) = 0,0025 mol ÷ 0,25 L.
c(KIO₃) = 0,01 mol/L = 0,01 M.

Ok where is the image ?