Answer to Question #193365 in Inorganic Chemistry for Strawberry Joy De

Question

0. 6.00 M of sulfuric acid, H2SO4(aq), has a density of 1.338 g/mL. (a) What is the percent by mass of 

sulfuric acid in this solution? (b) How to prepare a 100.0 mL of 1.25 M H2SO4 using this as a standard 

solution?

Answer 1

6.00 M is 6.00 moles/L of soln.

That is 6 moles x (98 g/mol) = 588 grams

H2SO4 in the liter of soln..

Take 1 L of the solution.

That will have a mass of volume x density = 1000 mL x 1.338 g/mL = 1338 grams.

How much is H2SO4 and how much sulfuric acid. There are 588 grams H2SO4; therefore, 1338 g - 588 = 750 g .

molality = mols/kg solvent.

Molarity = $\frac{ 6}{0.750}=8$