Question

. Al2(SO4)3 solution of 1 molal concentration is present in 1 litre solution of density 2.684
g/cc. How many moles of BaSO, would be precipitated on adding excess of BaCl, in it?
(a) 2 moles
(b) 3 moles
(c) 6 moles
(d) 12 moles
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Answer 1

Answer:

6 moles

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Answer 2

Answer:

Explanation:

Let x be the mass of al2(so4)3 in the solution .

Al2(so4)3+3BaCl2 ----------2AlCl3+3Baso4 equation number 1

1m =no of moles ÷ mass of solvent

No of moles = x÷ 342 here 342 is the molar mass

Mass of solvent = density ×volume-mass of al2(so4)3

Molar Mass of al2(so4)3=342

1 m=x÷342/(2684-x)×10^-3

On solving we get x =684

So number of moles of al2(so4)3 =684 / molar mass

684÷342=2 moles

As per the equation equation number 1

1 mole of al2(so4)3 gives 3 moles of baso4 precipitate

2 moles of al2(so4)3 gives 6 moles of baso4

So answer is (c)