Question

Pre-Lab Questions - Lab 9 When a student was completing this experiment, the atmospheric pressure was measured to be 751 torr. The temperature of the water was found to be 23.0 "C. The volume of water displaced (equal to the volume of oxygen produced) was found to be 325 mL. re? 1. What is the water vapor pressure at this temperature? 2. What is the pressure of the oxygen gas, Poz, that is produced? Record in units of atmospheres (atm) 3. Based upon the data provided, calculate the number of moles of O, formed during the reaction. 4. If the oxygen gas formed from this reaction was collected at at temperature of 30.0°C instead of 23.0 °C, explain how this might affect the data collected.

Answer 1

1 Atmospheric pressure = 751 torr

The pressure of water vapour at 23.0°C = 21.1 torr

2. Therefore the pressure of the oxygen gas

= total pressure – pressure of water vapour

= 751 – 21.1 = 729.9 torr.

760 torr = 1.0 atm

729.9 torr *1.00 atm / 760 torr

= 0.960 atm

3.

Now calculate the number of moles of gas by ideal gas law as follows:

PV= n RT

P = pressure, 0.960 atm

T = 23 C or 296 K

V is equal to the volume which is displaced by water = 325 ml or 0.325 L

Number of moles is calculated as follows:

n = PV/ RT

       = 0.960 atm * 0.325 L / 0.0821 atm. L /mole-K * 296 K

= 0.0128 moles   

4.If the temperature is 30 C therefore ;

The pressure of water vapour at 30.0°C = 31.8 torr

Therefore the pressure of the oxygen gas

= total pressure – pressure of water vapour

= 751 – 31.8 = 719.2 torr.

760 torr = 1.0 atm

719.2 torr *1.00 atm / 760 torr

= 0.946 atm