Question

need help to find answers for the blanks

Name Chem 2115 Experiment Date Section GAS BEHAVIOR Trial 1 I. Data Mass of test tube and contents 58.726 g 59.601 g before heating 58.619 g 58.726 g afterheating Masa loss 231.8 ml 259.6 mL 25.3 deg C 25.6 deg C 29.79 in Hg Volume of Co, collected, V Temperature of Co,,T Atmospheric pressure Water vapor pressure at Co, temperature 29.80 in Hg IL Calculations Mass of Co, collected Moles of Co,, System temperature (K),T Pressure of CO, (atm) Experimental R = pV/nT Molar Volume Vm = V/n Standard molar volume at STP VII-5

Answer 1

1. Mass loss = Mass before heating - Mass after heating = Mass of CO2 collected

Trial 1: Mass loss = 59.601 - 58.726 = 0.875 g

Trial 2: Mass loss = 58.726 - 58.619 = 0.107 g

2. Water vapour pressure at CO2 temperature:

Trial 1: Water vapour pressure at 25.3 oC temperature: 0.03182 atm

Trial 2: Water vapour pressure at 25.6 oC temperature: 0.0324 atm

Calculations	Trial 1	Trial 2
Mass of CO2 collected	0.875	0.107
Moles of CO2, n	0.0199 mol	0.00243 mol
System temperature (K), T	298.3	298.6
Pressure of CO2 (atm), p	0.996	0.9956
Volume of CO2 (L)	0.2596	0.2318
Experimental R = pV/nT	0.0435 L atm K-1 mol-1	0.3180 L atm K-1 mol-1
Molar volume, Vm = V/n	13.045 L/mol	96.40 L/mol
Standard molar volume at STP	12.985 L/mol	94.815 L/mol

3. Moles of CO2 = mass of CO2 collected/ molar mass of CO2

Trial 1: Moles of CO2 = 0.875/ 44 = 0.0199 mol

Trial 2: Moles of CO2 = 0.107/ 44 = 0.00243 mol

4. Pressure of CO2 , p = Atmospheric pressure - water vapour pressure

Trial 1: Pressure of CO2, p = 29.80 in Hg = (29.80* 25.4 mm Hg/ 760) = 0.996 atm

Trial 2: Pressure of CO2, p = 29.79 in Hg = (29.79* 25.4/ 760) = 0.9956 atm

5.

Trial 1: R = pV/ nT = (0.996 atm) (0.2596 L)/ (0.0199 mol) (298.3 K) = 0.0435 L atm K-1 mol-1

Trial 2: R = pV/ nT = (0.9956 atm) (0.2318 L)/ (0.00243 mol) (298.6 K) = 0.3180 L atm K-1 mol-1

6. molar volume, Vm = V/n

Trial 1: molar volume = 0.2596 L/ 0.0199 mol = 13.045 L/mol

Trial 2: molar volume = 0.2318 L/ 0.00243 mol = 96.40 L/mol

7. To calculate Standard molar volume at STP we will use the formula

$\frac{p_1V_1}{T_1} = \frac{p_2V_2}{T_2}$

p1, V1, and T1 = initial (experimental) pressure, volume and temperature

p2 and T2 = final pressure and temperature , i.e. STP

V2 = final volume at STP = ?

Trial 1: $V_2 = \frac{p_1V_1T_2}{p_2T_1} = \frac {(0.996 atm)(0.2596 L)(298.15 K)}{(1 atm)(298.3 K)}=0.2584\,L$

Trial 2: $V_2 = \frac{p_1V_1T_2}{p_2T_1} = \frac {(0.9956 atm)(0.2318 L)(298.15 K)}{(1 atm)(298.6 K)}=0.2304\, L$

These are the volumes of the gases reduced to STP.

Molar volumes are:

Trial 1: $V_2 = \frac{0.2584\,L}{0.0199 mol} = 12.985 \,L/mol$

Trial 2: $V_2 = \frac{0.2304\,L}{0.00243 mol} = 94.815 \,L/mol$

The actual value of standard molar volume at STP for an ideal gas = 22.4 L