100ml of 0.1M of Al2(SO4)3 is mixed with 100ml of 0.1M AlCl3. Molarity of Al+3 ion in the resulting solution is ____?

No. of molesn=MxV(ml)1000nAl2SO43=100X0.11000=0.01 mol and nAlCl3=100X0.11000=0.01 mol Total No. of moles of Al+3=0.02 mole  from Al2SO43)+0.01 mole  from AlCl3=0.03 mole ∴ Molarity of Al+3=nAl+3 Total Volume (ml)×1000=0.03200x1000=0.15M