Trial 1 Trial 2 Mass of empty tube after initial heating 5.448 .15.44log lu Mass

Question

Trial 1 Trial 2 Mass of empty tube after initial heating 5.448 .15.44log lu Mass of tube plus ple before heating (7,123 Mass of tube plus sample after heating (CoS Volume of water displaced Water temperature (C) Barometric pressure CALCULATIO Show your calculations in the space provided on the next page. Molar Volume of oxygen: Mass of oxygen generated Moles of O, generate Vapor pressure of water Pressure of O Temperature o Calculated volume of O, sample at STP Calculated lar volume of O, at STP mol Average molar volume Percent error in molar volume

Explanation / Answer

DATA SUPPLIED

Sr. No.

Trial 1

Trail 2

1

Mass of empty tube after initial heating

15.448 g

15.446 g

2

Mass of tube plus sample before heating

17.123 g

17.026 g

3

Mass of tube plus sample after heating

17.068 g

16.884 g

4

Volume of water displaced

164 mL

165 mL

5

Water temperature (C)

22.5

22.0

6

Barometric pressure

745.71 torr

CALCULATIONS

Sr. No.

Trial 1

Trail 2

7

Mass of oxygen generated (sr. no. 2 -3)

0.055 g

0.142 g

8

Moles of oxygen generated (check calculation below)

6.453*10-3 mole

6.515*10-3

9

Vapor pressure of water

21.1 torr (The value is obtained from vapor pressure table; the value is given at 23C and we have approximated the value to stay the same)

19.8 torr

10

Pressure of O2 (sr. no. 6 – sr. no. 9)

(745.71 – 21.1) = 724.61 torr

(745.71 – 19.8) = 725.91 torr

11

Temperature of O2 (K) (sr. no. 5 + 273)

295.5

295

12

Calculated volume of O2 sample at STP (check calculation below)

0.1444 L

0.1458 L

13

Calculated molar volume of O2 at STP (check calculation below)

22.315 L

22.379 L

14

Average Molar volume (check calculation below)

22.347 L

15

Percent error in molar volume (check calculation below)

0.24

Calculation for no. 8:

Pressure of O2 = 724.61 torr = (724.61 torr)*(1 atm/760 torr) = 0.9534 atm (1 atm = 760 torr)

Temperature of O2 = 295.5 K

Volume of O2 gas = volume of water displaced = 164 mL = (164 mL)*(1 L/1000 mL) = 0.164 L

We know that

n = PV/RT = (0.9534 atm)*(0.164 L)/(0.082 L-atm/mol.K).(295.5 K) = 6.4528*10-3 mole 6.453*10-3 mole.

Calculation for sr. no. 11:

The number of moles of O2 will stay constant; therefore, we can write

P1V1/T1 = P2V2/T2

Given P2 = 1atm and T2 = 273 K, V2 = (P1V1/T1)*(T2/P2)

Put P1 = 0.9534 atm, V1 = 0.164 L and T1 = 295.5 K, we have

V2 = (0.9534 atm)*(0.164 L)/(295.5 K)*(273 K/1 atm) = 0.1444 L

Calculation for sr. no. 12:

Moles of O2 = 6.453*10-3; volume occupied at STP = 0.1444 L.

Therefore, volume occupied by 1 mole of O2 at STP = (0.1444 L)*(1 mole/6.453*10-3 mole) = 22.315 L.

Calculation for sr. no. 14:

Average molar volume = ½*(22.315 + 22.379) L = 22.347 L

Calculation for sr. no. 15:

Accepted value of molar volume = 22.40 L; obtained value = 22.347 L.

Percent error = (accepted value – obtained value)/(accepted value)*100 = (22.40 – 22.347)/(22.40)*100 = 0.2366 0.24%

Sr. No.

Trial 1

Trail 2

1

Mass of empty tube after initial heating

15.448 g

15.446 g

2

Mass of tube plus sample before heating

17.123 g

17.026 g

3

Mass of tube plus sample after heating

17.068 g

16.884 g

4

Volume of water displaced

164 mL

165 mL

5

Water temperature (C)

22.5

22.0

6

Barometric pressure

745.71 torr

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