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

Let no. of moles of Al₂(SO₄)₃ in solution is x

$\frac{{\mathrm{n}}_{\mathrm{solute}}}{\mathrm{wt}. \mathrm{of} \mathrm{solvent}}\times 1000=1$
$\frac{\mathrm{x}}{2684-342\mathrm{x}}\times 1000=1$
 x= 2

${\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3+3{\mathrm{BaCl}}_2\to 3{\mathrm{BaSO}}_4+2{\mathrm{AlCl}}_3$
$\begin{array}{l}2{\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3⟶6 \mathrm{moles} {\mathrm{SO}}_4^{2-}\\ \therefore 6 \mathrm{moles} \mathrm{of} {\mathrm{BaSO}}_4\end{array}$