Please pro vide the answer using my calculations provided as well as sh ow the w

Question

Please provide the answer using my calculations provided as well as show the work and written steps on how to find the answer for questions 1-8 and questions a-c below. Thank you!

Calculations:

1.   Mercury equivalent of H2O column (calculation needed)   _______________

2.   Partial pressure of hydrogen gas (calculation)       _______________

3.   Mole of H2 collected (calculation)               _______________

4.   Coulombs of electricity used (calculation)           _______________

5.   Coulombs used per mole of H2 (g) (calculation)       _______________

6.   Electrons per mole (calculation)               _______________

7.   % error (calculation)                       _______________

8.   Mass of copper lost at anode per mole of electrons (calculation) ___________

Please provide THREE sources of error to your value of Avogadro's Number and explain how the error would affect my value using my data; I have already found 3 errors (a-c) so please explain how these would affect my data. Thank you!

Questions:

   Discuss three sources of error to your value of Avogadro's Number. Show how each would affect your value.

a.   Amps were not kept at 0.5 amps.

b.   Height measurement of water column was greater.

c.   H2 gas escapes.

•   Any other errors are fine to use as a-c are my examples of possible errors.

AVOGADRO'S NUMBER DETERMINATION BY ELECTROLYSIS

Background and Theory
   Since the mole concept is so important to an understanding of basic chemistry it is worthwhile to consider it in some detail. In this experiment you will measure the number of particles in a mole, this is usually called Avogadro's Number. By utilizing an electrolysis reaction we can measure the current and time for the reaction. From this the coulombs of electricity used can be calculated. If the moles of product formed at the cathode is measured a simple calculation will yield a value for coulombs per mole, dividing this by the electron charge will yield the electrons per mole or Avogadro's Number.

   This experiment involves the electrolysis of a dilute sulfuric acid solution using copper electrodes. The volume of hydrogen produced on the cathode for a given time and current is measured and the atmospheric pressure, temperature and height of the water column are recorded. In addition, if the mass decrease of the anode is measured, copper is oxidized to copper (II) ion, the equivalent weight of copper can be determined.

Procedure:
   You will find the apparatus set up in the laboratory. Obtain a copper anode from the side table and weigh it on the analytical balance. Fill the 400 mL beaker about half-full with distilled water and add 50 mL of dilute (6N) sulfuric acid. With the aspirator raise the water level in the buret until it is above the 50 mL mark. Using the pinch clamp adjust the level to the 50 mL mark.

   Insert the copper anode into the solution and connect the electrodes to the proper lead from the power supply. Note the time and turn the power supply on. Adjust the current to about 0.5 amp with the knob on the power supply. During the electrolysis be careful not to move the electrodes, since this would change the current. Allow the electrolysis to proceed until between 35 and 40 mL of hydrogen have been collected. Turn off the power supply and record the time. Measure the height of the water column, gas volume, temperature of the solution and barometric pressure. Use care when reading the buret since the scale will be inverted.

   Remove the copper anode, rinse in distilled water and then in acetone. Air dry and weigh on the analytical balance. If no one is waiting to use the apparatus empty the sulfuric acid solution and rinse with distilled water.
Data:
   Time electrolysis begins       _____0.00________
  
   Time electrolysis ends           ____7:18 min______
  
   Elapsed time               _7:18 min = 438 sec_
  

   Current during electrolysis:        __0.5 amp________

   Initial buret reading           __ 48.3 mL______
  
   Final buret reading           ____12.8 mL______
  
   Volume of hydrogen collected   ____35.5 mL______
  
   Height of water column       19.1 cm = 191.0 mm_

   Temperature of solution       ____ 22 C______
  
   Barometric pressure           ___30.10 in Hg ___
  
   Vapor pressure due to water       ___20.946 torr.____

  
   Initial mass of Cu anode       __ _4.2526 g_______
  
   Final mass of Cu anode         ____4.1625 g______

Explanation / Answer

1. Mercury equivalent of H2O column:The density of Hg is 13.6 times that of water.

Therefore, dHgxhhg = dWater xhwater density of water = 1.0 g/cc and that of Hg = 13.6 g /cc

hHg= dWater xhwater / dHg = 1 x 191/13.6 = 14.04 mm

2. Partial pressure of hydrogen gas:

Volume of H gas collected = 35.5 ml

WE need to calculate Volume understandard conditions P1V1/T1 = P2V2/T2

Temperature has remained constant. at 2 2 degree celsius = 273+22 = 295 K

Barometric pressure = 30.10 in = 30.10x25.4 = 764.54 mm

Standard molar volume of any gas = 22.4L at 273K and 760 mm pressure

ThereforeV1 x760/273 = 35.5 x 764.54 /295

V = 33.04ml

moles of hydrogen = 33.04/22400 = 1.47x10-3

PV = nRT

pressure exerted by the gas P = 1.47x10-3 mol x 0.0821 L atm K-1 mol-1x295 /  0.0355L = 1.002atm = 762.2 mm

This pressure is the total pressure exerted by hydrogen and water vapour (aqueous tension)

Hence partial pressure of Hydrogen = 762.2- 20.946 = 741.25 mm

3. Mole of H2 collected = moles of hydrogen = 33.04/22400 = 1.47x10-3

4. Coulombs of electricity used Q = I (in A) x t(in s) = 0.5 x 430.8 = 215.4C

5. Coulombs used per mole of H2 (g) 2H+  + 2 e -------- H2(g)

hence for 1 mol of H2 two moles of electrons are required

96500 C is the total charge carried by 1 mol of electrons

Hence coulombs used pe(s)r mole of hydrogen = 2x 96500 = 193000C

6.  Electrons per mole : Shown above one mol of hydrogen gas requires two moles of electrons

8 . Mass of copper lost at anode per mole of electrons

Cu(s) -------- Cu+2 + 2e- (molar mass of copper = 63.5)

two moles of electrons correspond to the loss of one mol of copper, that is 63.5g

therefore, for 1 mol of electrons 31.75 g are lost

% error Theoretically mass of copper lost = 215.4x31.75/96500 = 0.0708 g

in this experiment mass lost = 0.0901g

% error = 21.4%

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