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 0^oC 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

Answer 1

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/V_m where M = molecular weight of the gas (in g/mol) and V_m = molar volume of the gas (in L/mol). Consequently, we can write the ideal gas law as

P*V_m = R*T

====> V_m = 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 R_U. It is given that R_U = ½*R_O2 where R_O2 is the rate of diffusion of O₂.

Therefore, R_O2/R_U = R_O2/(1/2*R_O2) = 2.

The molecular weight of O₂ = 2*15.9994 = 31.9988.

Use Graham’s law as shown above.

R_O2/R_U = √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).