23 August, 03:20
How many grams of water must be added to 455 grams of potassium sulfate in order to make a 1.50 m solution?
23 August, 04:39
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.
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