Experiment3 PRE-LAB Questions (show instructor before starting, include in lab r

Question

Experiment3 PRE-LAB Questions (show instructor before starting, include in lab report) The hydrogen peroxide solution that you are using in this experiment is labeled as a 3% solution. mass/volume. However, in order to complete the calculations, the concentration must be in molarity. Calculate the molarity of a 3% mass volume 110 solution (Part I, IL and IV) and a 1.5% mass/volume H2O2 solution (Part III) and record these values in the table below 1. Complete the following table. NOTE: 3% mass/volume H2O2 solution means 3g H202 100ml of solution and you need to convert it to moles of H202/ L of solution. volume H02[:O] in Molarity Volume KI before mixing Part KI) before mixing 0.50 M 0.25 M 0.50 M 0.50 M Trial) (mL) 4of3% 4013% 40f 1.5% 4 of 3% mL) 2. Reading the table for Trial 1, what would be the new concentration of [KIl after mixing 4 ml of the H202 with 1 ml of the [KI)? What is the concentration of [I-1 in the same trial? (HINT sethedilution equation MiVi-Maw-son . lal=A.,L krK+ 3. Considering trial III fill in the blanks volume of 5 ml. What is the new concentration of H:02 after mixing from trial IlI? 4. You will need to do this for each Trial to complete the H:0:] after mixing and [I-] after I which will give me a total mixing' columns in the data table for the Postlab/Data analysis section.

Explanation / Answer

Answer:

Molarity of 3% (m/v) H2O2 solution

3 g H2O2 in 100 ml water

moles of H2O2 = 3 g/34 g/mol

                         = 0.088 mol

molarity = moles/L of solution

So,

molarity of 3% H2O2 solution = 0.088 mol/0.1 L

                                                = 0.88 M

Molarity of 1.5% (m/v) H2O2 solution

1.5 g H2O2 in 100 ml water

moles of H2O2 = 1.5 g/34 g/mol

                         = 0.044 mol

molarity = moles/L of solution

So,

molarity of 3% H2O2 solution = 0.044 mol/0.1 L

                                                = 0.44 M

  

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