Question

50 mL of a mixture consisting of 0.019 M cerium (IV) and 2.7 M H2SO4, or sulfuric acid. Begin by preparing 100 mL of 2.7 M H2SO4 using the available solution of 42.3% w/w H2SO4. This concentration unit, which may be less familiar to you, is a weight-to-weight percent (100.0 g of the solution contains 42.3 g of H2SO4). The density of 42.3% w/w H2SO4 is 1.25 g solution/mL solution. Using a graduated cylinder, measure out the correct volume of 42.3% w/w H2SO4 and slowly add it to a 100-mL volumetric flask that already contains approximately 25 mL of deionized water.

Answer 1

Answer:

The correct volume of 42.3% w/w H₂SO₄ is 50,1 mL

Explanation:

In order to know the correct volume of 42.3% w/w H₂SO₄ needed, we need to know the mass of H₂SO₄ that is needed. We'll use the wanted concentration, the final volume and the molecular weight of sulphuric acid, thus:

2,7 M * 0,1 L * 98,079 g/mol = 26,48 g of sulphuric acid are needed.

Now we can calculate the correct volume, using the concentration of the available solution, and the density of said solution, thus:

$26,48g H2SO4 *\frac{100g Solution}{42,3 g H2SO4}*\frac{1 mLsolution}{1,25gSolution}=50,08 mL$

We would then need to measure 50,1 mL (rounded up) of 42.3% w/w H₂SO₄ with the graduated cylinder, and then add it to the volumetric flask.