### CHAPTER 5 BASIC HOMEWORK

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**Question 1**

What is the pressure in atm of 383 mmHg? Help: Pressure–gases Answer:

**Explanation**

**Explanation**

1 atm = 760 mmHg 1/ 760 = X/ 383 X= 0.5039

**Question 2**

What is the pressure in torr of 1.47 atm? Answer:

**Explanation**

**Explanation**

760 torr = 1 atm 760 /1 = X / 1.47 X= 1117.2

**Question 3**

What is the pressure in atm of a gas where n = 0.16 mol, V = 13.76 L, and T = 28.4oC? Help: Ideal gas law Answer:

**Explanation**

**Explanation**

PV=nRT 28.4oC= 301.55 K R=0.08206 P(13.76) = (0.16)(0.08206)(301.55) P=0.2877

**Question 4**

What is the temperature in K of an ideal gas where P = 728 torr, n = 0.0271 mol, and V = 476.5 mL? Hint for finding T using PV = nRT Answer:

**Explanation**

**Explanation**

PV=nRT (728/760)(.4765)=(0.0271)(0.08206)T =205.24

**Question 5**

How many moles of an ideal gas produce a pressure of 744 torr and a volume of 2.30 L at 285 K? Answer:

**Explanation**

**Explanation**

PV=nRT (744/760)(2.30L) = n (0.08206)(285) n=0.0962744

**Question 6**

If the mass of the gas in the preceding problem is 33.1 g, what is the molar mass or molecular weight of the gas? Help: Molecular weight Answer:

**Explanation**

**Explanation**

33.1 g / X MM = 0.0962744 X= 343.81

**Question 7**

At a fixed temperature and number of moles, the initial volume and pressure of a helium gas sample are 238 mL and 685 torr, respectively. What is the final volume in mL, if the final pressure is 66.3 torr? Help: Boyle’s Law Answer:

**Explanation**

**Explanation**

P1V1 = P2V2 238(685) = 66.3 X X=2458.974

**Question 8**

At a fixed temperature and number of moles of nitrogen gas, its volume and pressure are 196 mL and 687 torr, respectively. What is the final pressure in torr, if the final volume is 290 mL? Answer:

**Explanation**

**Explanation**

196(687)= 290 X X=464.317

**Question 9**

At a fixed pressure and number of moles, the initial volume and temperature of an air sample are 491 mL and 270 K, respectively. What is the final volume in mL, if the final temperature is 481 K? Help: Charles Law Answer:

**Explanation**

**Explanation**

V1/T1 = V2/T2 491/270 = X/481 874.707407

**Question 10**

To what temperature in K must a variable volume container of helium gas be cooled to obtain a final volume of 0.255 L? The initial volume and temperature are 2.86 L and 332 K, respectively. The pressure and number of moles are fixed. Answer:

**Explanation**

**Explanation**

2.86/332 = 0.255 / X X=29.601

**Question 11**

A balloon filled with air and having no leak initially has a volume of 27.5 L, a pressure of 732 torr, and a temperature of 287 K. What is the final volume in L, if the final pressure is 489 torr and the final temperature is 244 K? Help: Ideal gas law, n fixed Answer:

**Explanation**

**Explanation**

P1V1/T1 = P2V2/T2 (732)(27.5)/287 = (489)T/244 T=34.997

**Question 12**

What is the total pressure if a container contains 59 torr of He gas, 234 torr of N2 gas, and 44 torr of Ar gas? Help: Dalton’s Law Answer:

**Explanation**

**Explanation**

59 + 234 + 44 = 337

**Question 13**

Oxygen gas is collected over water. The total pressure (the O2 pressure + the water vapor pressure) is 735 torr. The temperature of the water is such that the water vapor pressure is 44 torr. What is the partial pressure of the oxygen gas in torr? Answer:

**Explanation**

**Explanation**

735 – 44 = 691

**Question 14**

What volume in L of chlorine at 30oC and 702 torr is required to react completely with 4.48 g aluminum: 2Al(s) + 3Cl2(g) → 2AlCl3(s) Hint for chlorine Answer:

**Explanation**

**Explanation**

Procedure: g Al –> mol Al –> mol Cl2. Next solve for V in PV = nRT. Did you remember to convert torr to atm and C to K? 4.48 g Al / 26.981 MM = 0.166 mol Al 0.166 mol Al * 3 mol Cl2/ 2 mol Al = 0.24906 mol Cl2 PV= nRT (702/760)V= (0.24906)(0.08206)(303.15) V = 6.54