CHE 169 LC Exp. 8 MOLAR VOLUME OF A GAS Name hiidst No. Section Desk LABORATORY

Question

CHE 169 LC Exp. 8 MOLAR VOLUME OF A GAS Name hiidst No. Section Desk LABORATORY REPORT Date DATA: RECORD ALL DATA IN INK using the correct number of significant figures and correct units. 1. Mass of magnesium sample 2. Volume of hydrogen 3. Temperature of hydrogen 26.0 4. Temperature of water 766 5. Barometric pressure 6. Difference in water levels 7. Aqueous vapor pressure CALCULATIONS: SHOW COMPLETE SET-UPS" (and the rounding off FOR ALL CALCULATIONS on the back of this page or on a separate sheet. Report results to the correct number of significant figures and correct units. 1. Absolute temperature of the gas 2. Mercury equivalent of the difference of water levels 12.3 mm Ha 3. Partial pressure of dry hydrogen 4. Volume of dry hydrogen at STP 5. Moles of hydrogen 6. Molar volume at STP 7. Average molar volume at STP Average deviation 8. Dev. of average STP molar w from ideal STP molar V

Explanation / Answer

3)

The partial Pressure of hydrogen gas (trial-1) = 766.2 mmHg – 12.3 mmHg = 753.9 mmHg

The partial pressure for hydrogen gas (trial-2) 766.2mmHg - 13 mmHg is = 753.2 mmHg.

4)

Now, we can use the combined gas law to find the volume of hydrogen gas at STP.

The combined gas law is:

P1V1/T1 = P2V2/T2

Temperature is kept constant, so we can take out it from the equation.

       P1V1/p2 = V2

The known values, for Trial 1

P1 = 753.9 mmHg

P2 = 760.0 mmHg (this is the standard pressure in STP in mmHg units)

V1 = 81.0 ml = 0.081 L

For Trial 2

P1 = 753.9 mmHg

P2 = 760.0 mmHg

V1 = 82 ml = 0.082 L

Substitute them into the equation,

For Trial 1-

V2 = 753.9 mmHg • 0.081 L/760.0mmHg = 0.0803 L   

For Trial 2-

V2 = 753.2 mmHg • 0.082 L / 760.0 mmHg = 0.0812

5)

For Trial 1-

In order to get the moles of hydrogen gas, we can calculate the moles of magnesium, and then using the mole ratio from the equation, get the moles for hydrogen. So,

0.0777 g / 24.31 g/mol = 0.00319 mol of Mg

The mole ratio of one Mg per one H2 1: 1

Therefore, the theoretical moles of H2 in this experiment is 0.00319 mol of H2

Now, the volume will be divided by the number of moles of hydrogen gas to get the molar volume

0.0803 L H2/0.00319 mol H2 = 25.172L/mol H2

For Trial 2-

0.0769g /24.31 g/mol = 0.00316 mol of Mg

So, moles of H2 = 0.00316

6)

0.0812 L H2 / 0.00316 mol H2 = 25.696 L/mol H2

7)

Average molar volume at STP = (25.172L /mol + 25.694L/mol)/2 = 25.434L/ mol

Compare to the actual molar volume of hydrogen at STP, 22.4 L/mol   

22.4 L/mol H2 – 25.434 L/mol H2 = - 3.034 L/mol

Average deviation = 3.034/22.4 = 0.135

% deviation = 0.135 x 100 = 13.5

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---