A 0.0020 m aqueous solution in an ionic compound Co(NH3)5(NO2)Cl freezes at –0.007320C. Number of moles of ions which 1 mol of ionic compound produces on being dissolved in water will be (Kf = +1.86oC/m)

molality of Co(NH3)5(NO2)Cl = 0.002mFP of sol. = -0.00732ocKf of water = 1.86KKgmole-1No.of particles  produced by one molecule (or) No.of moles of ions produced by one mole of  compound (n) =?0.00732 = i(1.86)(0.002)7.32X10-3 = i(1.86)(2X10-3)Assuming 100% ionization, i=n=2The complex can be [Co(NH3)5(NO2)]Cl (or) [Co(NH3)5Cl]NO2