Given,molality = 0.0020 mΔ Tf = 0o C -0.007320 Ckf = 1.86 oC/mΔTf = i.kf x mi = ΔTf/ kf x m= 0.00732/1.82 x 0.0020= 1.92 = 2.....A 0.002M aqueous solution of an ionic compound [Co(NH3)5(NO2)]Cl freezes at -0.00732∘C. Find the number of moles of ions which 1 mole of ionic compound produces of being dissolved in water. (Kf=-1.86∘C/m).