Verified answer

delta t = m x Kf x i

here i = 3 ...3 particles in solution

so m = 2.5 / ( 1.86 x 3 ) = 0.448 moles of solute per kg of solvent or 0.448 x 142.042 g

= 63.63 g / Kg or 63.63 x 75 / 1000 g in 75 g of water

= 4.77 g of Na2SO4 to lower the freezing point 2.5 oC

delta t = m x Kf x i

here i = 3 ...3 particles in solution

so m = 2.5 / ( 1.86 x 3 ) = 0.448 moles of solute per kg of solvent or 0.448 x 142.042 g

= 63.63 g / Kg or 63.63 x 75 / 1000 g in 75 g of water

= 4.77 g of Na2SO4 to lower the freezing point 2.5 oC

after the first step, you take advantage of that answer you acquire for molality and plug it into the definition of molality so m=mol solute/ kg solute bear in mind that mols is an similar as grams/molecular weight, locate the molecular weight of ethylene glycol and sparkling up for grams to get your answer