How many grams of water must be added to 455 grams of potassium sulfate in order to make a 1.50 m solution?

Molar mass of K₂SO₄ = 174.3 g/mol
Mass of K₂SO₄ = 455 g

Moles = mass / molar mass

Hence, moles of K₂SO₄ = 455 g / 174.3 g/mol
                                      = 2.61 mol

Molarity = moles of solute (mol) / Volume of the solution (L)

Hence, volume = moles of solute / molarity
                         = 2.61 mol / 1.50 M
                         = 1.742 L = 1742 mL

Hence, needed volume of water = 1742 mL

Density = mass / volume

Density of water = 0.9972 g/mL

Hence mass of water = 0.9972 g/mL x 1742 mL
                                   = 1737.12 g

Hence, needed mass of water = 1737.12 g.

Answer) by studying proton collisions hope this : )