How many grams of lithium sulfate are required to make 2500 mL of a 2.67 M solution of lithium sulfate? *

A. 248 g

B. 734 g

C. 527 g

D. 812 g

The correct answer is B. 734 g

Explanation:

The chemical formulae of lithium sulfate is Li₂SO₄. With this, we can calculate the molecular weight (MM) of lithium sulfate as follows:

MM (Li₂SO₄) = (2 x molar mass Li) + molar mass S + (4 x molar mass O)

= (2 x 6.9 g/mol) + 32 g/mol + (4 x 16 g/mol)

= 109.9 g/mol

We need to prepare a solution with a molarity of 2.67 M. That means that the solution has to have 2.67 moles of Li₂SO₄ per liter of solution. We can convert from mol to grams with the calculated molecular weight and then we have to multiply by the volume of solution (2500 ml = 2.5 L), as follows:

Mass = 2.67 mol/L x 109.9 g/mol x 2.5 L ≅ 734 g