1. Calculate the molecular weight of a gas if 600 ml of the gas measured at 30 o
ID: 547538 • Letter: 1
Question
1. Calculate the molecular weight of a gas if 600 ml of the gas measured at 30 oC and 630 torr has a mass of 0.60 g.
a. 30 b. 40 c. 50 d. 60
2. Calculate the density , in g/ li , of oxygen gas at 25 o C and 1.05 atm pressure.
a. 3.37 b. 2.37 c. 1.37 d. 0.37
3. If the 1.2 moles of a gas are confined in a 10 liter container at a temperature of 100oC, what is the pressure, in atm?
a. 1.7 b. 2.7 c. 3.7 d. 4.7
4. If a gas diffuses at a rate of 1/ 2 as fast as oxygen gas, what is the molecular weight of the gas?
a. 128 b. 228 c. 328 d. 428
6. A sample of gas occupies a volume of 225 ml. at a pressure of 720 torr and a temperature of 20 o C. Calculate the new pressure in torr, if the volume is increased to 350 ml, at constant temperature.
a. 263 b. 363 c. 463 d. 563
7. A sample of gas occupies a volume of 275 ml at 20 oC and 1 atm pressure. Calculate the volume in ml. of the gas at 0oC and 1 atm pressure.
a. 256 b. 356 c. 456 d. 556
8. Calculate the total pressure, in torr, of a mixture of gases, each of which exerts the given partial pressure: H2, 150 torr ; N2, 210 torr; He, 320 torr.
a. 580 b 680 c. 780 d. 880
9. A sample of gas occupies a volume of 512 ml at 20 oC and 740 torr. What volume in ml. would this gas occupy at STP.
a. 364 b. 464 c. 564 d. 664
10. A sample of oxygen gas collected over water occupies a volume of 210 ml at 22 oC and 750 torr pressure. Calculate the volume in ml of dry oxygen at STP.
a. 487 b. 387 c. 287 d. 187
Explanation / Answer
1) The pressure of the gas, P = 630 torr = (630 torr)*(1 atm/760 torr) = 0.82895 atm; the volume of the gas is V = 600 mL = (600 mL)*(1 L/1000 mL) = 0.600 L; the temperature of the gas is T = 30°C = (273.15 + 30) K = 303.15 K.
Use the ideal gas law: P*V = n*R*T where n = number of mole(s) of the gas = (mass of the gas)/(molecular weight of the gas) = (0.60 g/MW) where MW = molecular weight of the gas.
(0.82895 atm)*(0.600 L) = (0.60 g/MW)*(0.082 L-atm/mol.K)*(303.15 K)
====> 0.49737 = (14.91498/MW) g/mol
====> MW = (14.91498/0.49737) g/mol = 29.9876 g/mol 30.0 g/mol.
The MW of the gas can be expressed as a dimensionless quantity; hence the correct answer is (a) 30.
2) The pressure of the gas is P = 1.05 atm while the temperature of the gas is T = 25°C = (25 + 273.15) K = 298.15 K.
Let d be the density of the gas. We know that d = M/Vm where M = molecular weight of the gas (in g/mol) and Vm = molar volume of the gas (in L/mol). Consequently, we can write the ideal gas law as
P*Vm = R*T
====> Vm = R*T/P
====> M/d = R*T/P
====> d = P*M/RT
The molecular weight of oxygen (in g/mol) = (2*159994) g/mol = 31.9988 g/mol.
Plug in values now.
d = (1.05 atm)*(31.9988 g/mol)/(0.082 L-atm/mol.K).(298.15 K) = 1.37428 g/L 1.37 g/L.
The correct answer is (c).
3) The volume of the gas is V = 10.0 L while the temperature of the gas is T = 100°C = (100 + 273.15) K = 373.15 K.
Use the ideal gas law:
P*V = n*R*T where n = number of moles of gas = 1.2 mole.
Plug in values.
P*(10.0 L) = (1.2 mole)*(0.082 L-atm/mol.K)*(373.15 K)
====> P = (1.2 mole)*(0.082 L-atm/mol.K)*(373.15 K)/(10.0 L) = 3.6718 atm 3.7 atm .
The correct answer is (c).
4) The rates of diffusion of two gases are related to the molecular masses of the gases as
Rate1/Rate2 = M2/M1 where M1 and M2 are the molecular weights of the two gases.
Let the unknown gas be denoted as U and the rate of diffusion of U is RU. It is given that RU = ½*RO2 where RO2 is the rate of diffusion of O2.
Therefore, RO2/RU = RO2/(1/2*RO2) = 2.
The molecular weight of O2 = 2*15.9994 = 31.9988.
Use Graham’s law as shown above.
RO2/RU = M/31.9988 where M = molecular weight of U.
===> 2 = M/31.9988
===> 4 = M/31.9988
===> M = 4*31.9988 = 127.9952 128
The correct answer is (a).
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