The container tells us that con H2SO4 (aq) is 96.0% by mass H2SO4 and has a density of 1.84 g/mL. In carrying out the calculation, assume you are starting with 1.00 L of con H2SO4 (aq).

Question

a) What is the mass of 1.00 L of con H2SO4 (aq)?

b) How many grams of H2SO4 are in this 1.00 L of solution? How many moles of H2SO4 is this?

c) Calculate the morality of this con H2SO4 solution.

d) How many liters of con H2SO4 (aq) must the stockroom personnel use to prepare 5.0 L of 1.75 M H2SO4?

Answer 1

a) Use the density to find this: 1.00 L x 1000 ml/L x 1.84 g/ml = 1840 grams

b) Since the solution is 96.0% H₂SO₄ by mass, we can find the mass of H₂SO₄.

1840 g x 0.960 = 1766 g H₂SO₄

Moles H₂SO₄ = 1766 g H₂SO₄ x 1 mole/98 g = 18.0 moles

c) I think you mean to calculate the molality (not morality...LOL):

molality (m) = moles H₂SO₄/kg solvent

We know there are 1766 g H₂SO₄ (18.0 moles), so to find the g or kg of solvent, we deduct the mass of H₂SO₄ from the total mass of solution: 1840 g - 1766 g = 74 g solvent = 0.074 kg solvent

Molality = 18.0 moles/0.074 kg = 243 m

d) V1M1 = V2M2

First, find the molarity of the original concentrated H₂SO₄:

18.0 moles H₂SO₄/1 L = 18.0 M

(x L)(18.0 M) = (5.0 L)(1.75 M)

x = 0.49 liters of the original concentrated H₂SO₄ will be needed