Question -
A solution of (Al2(SO4)3) contain 22% salt by weight and density of solution is 1.253.The molarity, normality and molality of the solution is
Options:
A .  0.806 M, 4.83 N, 0.825 m
B .  0.825 M, 48.3 N, 0.805 m
C .  4.83 M, 4.83 N, 4.83 m
D .  None
Answer: Option A : A Molecular wt. of Al2(SO4)3=2×27+3×(32+4×16)=342g/mole Equivalent wt. of Al2(SO4)3=Eq.wt.ofAl3++Eq.WtofSO2−4=273+962 =57geq No. of. equivalent per mole =34257=6 Let volume of solutions = 1 L Wt. of solutions = V×density=1000×1.253=1253g Wt. of solute = 1253×22%=257.66 Moles. of solute = 275.66342=0.806 Wt. of solvent = 1253−257.66=977.34 Molarity = molesofsoluteVolumeofsolution=0.806M Normality = 6×Molality=6×0.806=4.836N Molalit = MolesofsoluteWt.ofsolventink.g=0.806977.34/1000=0.825m
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