Question

Table 1. Experimental Data

Mass of Cylinder and solution: 35.1967g

Mass of 10 mL cylinder before filling: 25.7418g

Mass of 10 mL cylinder after emptying: 25.7528g

Volume of solution in 10 mL cylinder: 9.56mL

Volume of gas in 100mL cylinder: 92.1mL

Temperature: 19.5C

Atmospheric Pressure: 29.98 inHg

Q3. 1,°C 17.0 18.0 Use the table of water's vapor pressure shown below. What is the vapor pressure in the unit needed for the ideal Gas Law? Record the vapor pressure in Table 2. P.torr 1,°C P,torr 1,°C P,torr 1,°C P,torr 1,°C P,torr 1,°C P torr 14.6 19.0 16.5 21.0 18.7 23.0 21.1 25.0 23.8 27.0 26.7 15.5 20.0 17. 6 22. 0 19.8 24. 0 22.4 26.0 25.2 28.0 28.4

Answer 1

Temperature of collected gas (K) = (temperature in ºC) + 273.15

= 19.5 + 273.15

= 292.65

Vapor pressure of H2O in 10 mL cylinder (atm)

Since the temperature of the lab is 19.5ºC which is the mean of 19.0ºC and 20.0ºC, the vapor pressure of H2O can be approximated as the mean of the vapor pressures at 19.0ºC and 20.0ºC.

Therefore, vapor pressure of H2O at 19.5ºC = ½*(16.5 + 17.6) torr

= 17.05 torr

= (17.05 torr)*(1 atm)/(760 torr)

= 0.0224 atm

Pressure of dry O2 gas (atm)

Atmospheric pressure = 29.98 inHg

= (29.98 inHg)*(1 atm)/(29.92 in Hg) [ 1 atm = 29.92 inHg]

= 1.002 atm

Pressure of dry O2 gas = (atmospheric pressure) – (vapor pressure of water at 19.5ºC)

= (1.002 atm) – (0.0224 atm)

= 0.9996 atm

Volume of O2 gas (L) = (volume in mL)*(1 L)/(1000 mL)

= (92.1 mL)*(1 L)/(1000 mL)

= 0.0921 L

Moles of O2 gas

Use the ideal gas law.

PV = nRT

where n is the number of moles of O2 gas obtained.

Therefore,

n = PV/RT

Plug in values and get

n = (0.9996 atm)*(0.0921 L)/(0.082 L-atm/mol.K)(292.65 K)

= 0.003836 mol

≈ 0.00384 mol

Moles of H2O2 that reacted

Consider the decomposition of H2O2 to O2 as below.

2 H2O2 (aq) -----------> 2 H2O (l) + O2 (g)

As per the stoichiometric equation,

2 moles H2O2 = 1 mol O2.

Therefore,

Mols H2O2 reacted = (0.00384 mol O2)*(2 mols H2O2)/(1 mol O2)

= 0.00768 mol

Mass of H2O2

The atomic masses are

H: 1.008 g/mol

O: 15.999 g/mol

Gram molar mass of H2O2 = (2*1.008 + 1*15.999) g/mol

= 18.015 g/mol

Mass of H2O2 = (0.00768 mol H2O2)*(18.015 g/mol)/(1 mol)

= 0.138 g

Mass of H2O2 solution transferred to flask

Mass transferred = (mass of cylinder and solution) – (mass of cylinder before filling)

= (35.1967 g) – (25.7418 g)

= 9.4549 g

Percent H2O2 by mass in solution = (mass of H2O2)/(mass of H2O2 solution transferred to flask)*100

= (0.138 g)/(9.4549 g)*100

= 1.4596

≈ 1.50