Question

. How many grams of Na2SO4 are required to make 2700 mL of a 2.0 M solution?

Answer 1

Answer:

Explanation:

From the net ionic equation

Ba2+(aq) + SO42-(aq) ==> BaSO4(s) we see that 1 mole Ba2+ reacts with 1 mole SO42- to -> 1 mol BaSO4

Find moles of Ba2+ used: 0.250 moles/L x 0.0323 L = 0.008075 moles Ba2+

Find moles SO42- present: 0.008075 moles Ba2+ x 1 mol SO42-/1 mol Ba2+ = 0.008075 mol SO42-

Find mass of Na2SO4 present: 0.008075 mol SO42- x 1 mol Na2SO4/1 mol SO42- x 142.04 Na2SO4/mole = 1.14698 g = 1.15 g Na2SO4 (to 3 significant figures)

Answer 2

Answer: Approx 2kg of salt are required.

Explanation: If 4.5 ⋅ L of a 3.0 ⋅ m o l ⋅ L − 1 solution are required, this is a molar quantity of 4.5 ⋅ L × 3.0 ⋅ m o l ⋅ L − 1 = 13.5 ⋅ m o l . Now, we simply multiply this molar quantity by a molar mass to get the mass: 13.5 ⋅ m o l × 142.04 ⋅ g ⋅ m o l − 1 = ? ? ⋅ g