Question

A sulfuric acid solution has a density of 1.49 g/mL and contains 59.0% H2SO4 by mass.

(a) What is the molarity (M) of this solution? (b) What is the molality of this solution?

Answer 1

a) 8.9635 M H2SO4

b) 6.0158 m H2SO4

Step-by-step explanation:

a) Assume the volume to be 1 L (1000 mL).

In order to find the mass of the solution, multiply density by volume.

`(1.49 grams)/(1 mL) x 1000 mL` = 1490 grams of solution.

If 59% of the solution is H2SO4, then multiply the mass of the total solution by 59% (0.59).

0.59 x 1490 = 879.1 grams of H2SO4

Now, the grams of H2SO4 can be converted to moles through the molar mass of H2SO4 (I used 98.076 g/mol).

879.1 grams H2SO4 x
`(1 mol)/(98.076 grams)` = 8.9635 moles of H2SO4

To find molarity, divide moles by liters (the original 1 L we used).

8.9635 moles/1 L = 8.9635 M H2SO4

b) Molality is moles divided by kilograms.

We already calculated the solution to be 1490 grams. Now we must convert to kilograms by dividing grams by 1000.

1490/1000 = 1.49 kg of total solution

We already calculated moles of H2SO4 to be 8.9635.

8.9635/1.49 = 6.0158 m H2SO4