You need to make an aqueous solution of 0.182 M chromium(III) nitrate for an experiment in lab, using a 500 mL volumetric flask.?

No. of moles of Cr(NO₃)₃ = (0.182 mol/L) × (500/1000 L) = 0.0910 mol

Molar mass of Cr(NO₃)₃ = (52.0 + 14.0×3 + 16.0×9) g/mol = 238.0 g/mol
Mass of Cr(NO₃)₃ should be added = (238.0 g/mol) × (0.0910 mol) = 21.7 g