8.In the first experiment, the maximum volume of oxygen produced was 96cm3
measured atrit.p. Calculate the concentration of the aqueous hydrogen peroxide
in mol/dm3.2H2O2(aq) 2H200) + O2(g) D)number of moles of O2 formed =
1] ii)number of moles of H202 in
40cm3 of solution
[1] iii)concentration of
the aqueous hydrogen peroxide in mol/dm3 =
Answers
8.
The equation for the reaction is: 2H2O2----------> 2H2O + O2
Given, 96 cm3 of O2 was evolved at RTP.
Hence, number of moles of O2 formed = ?
RTP conditions are: 25 degrees Centigrade and 1 atm.
=> 25+273 =>298 K and 1 atm.
We know that, 1 mole of any gas at STP has a volume equal to 22,400 mL.
So, let us first convert 96 cm3 of O2 at NTP to STP.
V1 = 96 cm3 V2 = ?
T1 = 298 K T2 = 273 K
P1 = 1 atm P2 = 1 atm
By gas equation:
P1V1/T1 = P2V2/T2
1 * 96/298 = 1 * V2/273
V2 = 273*48/149
V2 = 87.94 cm3
Hence, at STP, the volume will be 87.94 cm3.
As 1 mole of gas at STP has a volume equal to 22,400 cm3.
So, 87.94 cm3 is 87.94/22,400 moles => 0.0039 moles of Oxygen gas.
∴ The number of moles of Oxygen gas evolved is 0.0039 moles.
2H2O2----------> 2H2O + O2
^From the equation it is visible that, 2 parts of Hydrogen peroxide gives rise to 1 part of Oxygen gas.
Hence, 1 part of Oxygen gas is 0.0039 moles of O2.
So, 2 parts of H2O2 is 2 x 0.0039 moles of H2O2 =>0.0078 moles H2O2
As, it is already given to us that 40 cm3 of H2O2 was used.
So, 40 cm3 of H2O2 solution contains 0.0078 moles of H2O2.
=> 40/1000 liters => 0.04 liters of H2O2 sol. has 0.0078 moles of H2O2.
1 dm3 = 1 liter
So, 0.04 dm3 = 0.04 liters.
Hence, 0.04 dm3 of H2O2 soln. contains 0.0078 moles of H2O2.
Concentration (in mol/dm3) = 0.0078/0.04 =>0.78/4=> 0.195 mol/dm3
Hence, the molar concentration of the H2O2 solution is 0.195 mol/dm3.