Question

How many grams of the electrolyte Al2(SO4)3 (molar mass=142 g/mole) are required to make 325 mL of solution having an osmotic pressure of 675 mm Hg at 25 oC?​

Answer 1

Answer: 0.36g

Explanation:

π = iMRT

π = (inRT)/V

π = (imRT)/(MrV)

When we make m the subject of the formula we have;

m = (πMrV)/(iRT)

m=(0.8882atm x 142g/mol x 325x10-3L)/(5 x 0.08206LatmK-1mol-1 x 298.15K)

m = 0.335078176662g = 0.36g