Please answer E AND F Analysis is one of the many things chemists do. In this ex

Question

Please answer E AND F Analysis is one of the many things chemists do. In this experiment, you 'ill analyze a sample for its Hydrogen Peroxide content. To do this you will use three different methods. Titration with Potassium Permanganate Catalytic Decomposition to Water and Oxygen Oxidation of Iodide to Iodine; Spectroscopic measurement of Trilodide ion. You weigh out Potassium Permanganate. Mass of KMnO_4 (MW = 158.034) ........ 0.1927 g You dissolve it completely in water ..... 41.00 ml. Molarity of MnO_4 solution You weigh out your UNKNOWN .... 1.10 ml. You acidify and titrate it with your KMnO_4 solution until the permanganate color persists. Volume of permanganate solution .... 15.68 mL Molarity of the HOOH in the UNKNOWN Your pipet your UNKNOWN into a reaction tube.... 4.20 ml. Add a "boat" containing Manganese Dioxide. MnO_2 Set up the gas collection apparatus shown on the right. You sink the boot and allow the oxygen generated to collect in the collection task. The water temperature is ..... 22 degree C The Barometric Pressure is 771.8 mm Kg Weight of the collection Mask with remaining water... 221.25 g. Weight of collection flask tilled with water 272.38 g. Density H_2O 0.996 g/cc Partial Pressure of Oxygen collected Volume of oxygen collected Millimoles of oxygen gas generated Molarity of Hydrogen Peroxide in UNKNOWN. Put into a clean 250 mL volumetric flask exactly 0.90 ml of 0.90 mM iodate STANDARD. Add an excess of 1 M K 1 and 1 M Sulfuric Acid and let stand for 5 minutes. Dilute to volume. UNKNOWN: Pipet into a clean 250 mL volumetric flask 1.00 mL of your UNKNOWN and dilute to volume with distilled water. Put into a dean 250 mL volumetric flask exactly 0.90 mL

Explanation / Answer

froom gas law, n= PV/RT

n= moles of oxygen generated, P= 751.973/760 atm,=0.989 atm 0 V= 51.33ml =51.33/1000=0.05133 L T= 22+273=295K

R=0.0821 L.atm/mole.K, n= 0.989*0.05133/(0.0821*298)= 0.002075 moles=2.074 millmoles of O2

from the reaction 2H2O2-------->2H2O+ O2

mole of H2O2 required = 0.002075*2= 0.00415 moles, ml added= 4.2 ml =4.2/1000L =0.0042L

Concentration of H2O2= 0.00415/0.0042 = 0.988M

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