Answer :
Answer:
0.25 L of a solution with a molarity of 6M has 6*0.25 = 1.5 moles of the solute. The molar mass of ammonium sulfate is 132.14 g/mole. The mass of 1.5 moles is 132.14*1.5 = 198.21 g. Therefore 198.12 g of ammonium sulfate are required to make 0.25 L of a solution with a concentration of 6M.
Explanation: