What is the molarity of a solution when 25.0 g of aluminum sulfite is dissolved in distilled water and made up to 175.0 ml of solution?

Al2(SO3)3 has r.f.m.= (2x27 + 3x32 + 9x16) = 294 Thus 25 g is 25/294 = 0.085 moles of aluminium sulfite.

We have 0.085 moles in 175 ml of solution, which would be 0.085 x 1000/175 moles in a litre

= 0.486 M