Question

What is the molarity of a solution containing 49.0 g of sulfuric acid, H2SO4,

in 3.00 liters of solution? Answer with 3 sig figs.

Answer 1

Molarity = moles of solute/volume of solution in liters.

Our solute here is sulfuric acid, of which we have 49.0 g. The number of moles of sulfuric acid is (49.0 g)/(98.079 g/mol) = 0.4996 mol H2SO4. To obtain the molarity of this solution of sulfuric acid, we divide 0.4996 mol H2SO4 by 3.00 L, which gives us 0.167 M H2SO4 (answer given to three sig figs).