Question

how many grams of K2SO4 molar mass=174 g/mol would be needed to prepare 4.00 L of a 0.0510 M solution? ...?

Answer 1

Molarity = moles of solute/ volume of solution. Moles= weight of substance / molecular weight of that substance. Therefore, 0.051= moles of K2SO4 / volume of solution. 0.051= weight of K2SO4 /molecular weight of K2SO4 x volume of solution. Therefore, weight of K2SO4 =0.051 x 174 / 4 So weight = 2.2185 gm

Answer 2

each litre should contain exactly 0.051 moles of of K2SO4 which should weigh 174*0.051

0.051*4*174 = 35.496g