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
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
Was this answer helpful ?
:
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
Was this answer helpful ?
Submit Solution