# Al2(SO4) 3 solution of 1 molal concentration is present is 1 litre solution of 2.684 gm/ cc. How many moles of BaSO4 would be precipitated on adding BaCl2 in excess.

Dear Student,

The density of the solution is 2.684 g/cc, which includes both the solute Al_{2}(SO_{4})_{3} and the solvent water. So, deducing the density of water which is ~ 1 g/cc, we will have 1.684 g/cc of Al_{2}(SO_{4})_{3} .

So, in a litre of solution we have 1.684 x 1000 = 1684 g of Al_{2}(SO_{4})_{3} .

Moles of Al_{2}(SO_{4})_{3} =

$\frac{1684g}{342gmo{l}^{-1}}=4.92mol[MolarmassofA{l}_{2}(S{O}_{4}{)}_{3}=342gmo{l}^{-1}]$

Chemical reaction: $A{l}_{2}(S{O}_{4}{)}_{3}+3BaC{l}_{2}\to 3BaS{O}_{4}+2AlC{l}_{3}$

For 1 mol of Al_{2}(SO_{4})_{3} we will have 3 mol of BaSO4, therefore, for 4.92 mol of Al_{2}(SO_{4})_{3} it will produce 3 x 4.92 = 14.76 mol of BaSO_{4}.

Best Wishes !

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