Homework, chapter 12: 11, 15, 19, 27, 35, 43, 50
11. A gas occupies a volume of 410 mL at 27 oC and 740 mm Hg pressure. Calculate the
volume the gas would occupy at STP.
 P1V1  =  P2V2   V2 =  P1V1T2  =  740 mm Hg x 410 mL x 273 K  = 363 mL
300 K x 760 mm Hg
 T1   T 2 
 P2T1  

15. A 775 mL sample of NO2 gas is at STP. If the volume changes to 615 mL and temperature to
25 oC, what will be the new pressure?
P1V1T2
1 atm x 775 mL x 298 K
P2 =  V T  =  273 K x 615 mm Hg  = 1.38 atm
 2 1  

19. A mixture contains H2 at 600. torr pressure, N2 at 200. torr pressure, and O2 at 300. torr
pressure. What is the total pressure of the gases in the system?
Ptot = P1 + P2 + P3 = 600. torr + 200. torr + 300. torr = 1100. torr
27. How many grams of NH3 are present in 725 mL of the gas at STP?
PV  
1 atm x 0.725 L
 = 0.0324 mol
PV = nRT  n = 
=
 RT   0.082 (L atm /mol K) x 273 K 
17.03 g NH3
0.0324 mol x  1 mol NH  = 0.552 g NH3.

3 
35. Calculate the density of the following:
38.00 g
1 mol
a) F2 gas at STP. d=  1 mol  x  22.4 L  = 1.70 g/L

 

o
b) F2 gas at 27 C and 1.00 atm. Calculate new volume:
P1V1T2   1 atm x 22.4 L x 300 K 
V2 = 
=
= 24.6 L
273 K x 71 atm
 P2T1  

38.00 g
1 mol
d=  1 mol  x  24.6 L  = 1.54 g/L

 

43. In the lab, students generated and collected H2 gas according to the following equation:
Zn(s) + H2SO4(aq)  H2(g) + ZnSO4(aq)
a) How many mL of H2 at STP were generated from 52.7 g of Zn?
1 mol Zn
1 mol H2
52.7 g Zn x  65.39 g Zn  x  1 mol Zn  = 0.806 mol H2.

 

nRT
0.806 mol x (0.082 L atm/mol K) x 273 K 
PV = nRT  V =  P  = 
= 18.0 L =
1 atm

 

1.80 x 104 mL H2.
50. Explain why is it necessary to add air to a car’s tires during winter.
The pressure and temperature are related: P is proportional to T if all other parameters stay
constnt. Thus, as the temperature lowers during winter months, pressure is lowered as well. To
maintain good traction with the road, the tires must be kept at a certain pressure, thus adding air
to increase the pressure is necessary.
Homework, chapter 14: 1, 3, 21, 27 (a-c), 33
1. Which of the substances listed below are reasonably soluble, and which are insoluble in
water?
a) KOH – all potassium compounds are soluble; b) NiCl2 – most chlorides are
soluble, c) ZnS – most sulfides are insoluble, d) AgC2H3O2 – all acetates are
soluble, e) Na2CrO4 – all sodium compounds are soluble.
3. Calculate the mass percent of the following solutions:
a) 15.0 g KCl + 100.0 g H2O. (15.0 g KCl / 115 g solution) x 100 = 13.0 % KCl
b) 2.50 g Na3PO4 + 10.0 g H2O. (2.50 g / 12.5 g solution) x 100 = 20 % Na3PO4
c) 0.20 mol NH4C2H3O2 + 125 g H2O.
 77.09 g NH4C2H3O2  = 15.4 g NH C H O
0.20 mol NH4C2H3O2 x 

4 2 3 2
 1 mol NH4C2H3O2 
(15.4 g NH4C2H3O2/ 140.4 g solution) x 100 = 11.0 % NH4C2H3O2.
d) 1.50 mol NaOH in 33.0 mol H2O.
 40.0 g NaOH  = 60.0 g NaOH
1.50 mol NaOH x 

 1 mol NaOH 
 18.02 g H2O  = 595 g H O.
33.0 mol H2O x 

2
 1 mol H2O 
(60.0 g NaOH / 655 g solution) x 100 = 9.16 % NaOH.
21. What will be the molarity of the resulting solutions made by mixing the following?
Assume that the volumes are additive.
a) 125 mL of 0.5 M H3PO4 with 775 mL of H2O. V1M1 = V2M2,
V1M1
125 mL x 0.5 M
M2 =  V  =  (775 + 125) mL  = 0.0694 M.
 2  

b) 250 mL of 0.25 M Na2SO4 with 750 mL of H2O.
V1M1
250 mL x 0.25 M
M2 =  V  =  (750 + 250) mL  = 0.0625 M.
 2  

c) 75 mL of 0.50 M HNO3 with 75 mL of 1.5 M HNO3.
0.075 mL x 0.50 mol/L = 0.0375 mol
0.075 mL x 1.5 mol/L = 0.1125 mol
total = 0.15 mol HNO3.
0.15 mol / 0.150 L = 1 mol/L = 1 M.
27. Use the equation to calculte the following:
3 Ca(NO3)2(aq) + 2 Na3PO4(aq)  Ca3(PO4)2(s) + 6 NaNO3(aq)
molar mass 94.97 po4 Na3PO4: 163.9 Ca3(PO4)2: 310.2
a) the moles Ca3(PO4)2 produced from 2.7 mol Na3PO4.
1 mol Ca3(PO4)2 
2.7 mol Na3PO4 x 
 = 1 4 mol Ca3(PO4)2
 2 mol Na3PO4 
b) the moles NaNO3 produced from 0.75 mol Ca(NO3)2.
6 mol NaNO3 
0.75 mol Ca(NO3)2 x 
 = 2.25 mol NaNO3.
 2 mol Na3PO4 
c) the moles Na3PO4 required to react with 1.45 L of 0.225 M Ca(NO3)2.
1.45 L x 0.225 mol Ca(NO3)2 / 1L = 0.326 mol Ca(NO3)2.
 2 mol Na3PO4  = 0.218 mol Na PO .

3
4
 3 mol Ca3(PO4)2 
33. What is the (a) molality, (b) freezing point and (c) boiling point of a solution
contaning 2.68g of naphthalene in 38.4 g benzene?
Molar mass: naphthalene: 128.2, benzene: 78.1
 2.68 g npht  x  1 mol npht  x  1000 g benz = 0.544 m.
a)
 38.4 g benz   128.2 g npht   1 kg benz 
c)
tb = 0.544 m x 2.53 (oC kg bnz/mol npht) = 1.38 oC;
tb = 80.1 + 1.38 = 81.4 oC.
b)
tf = 0.544 m x 5.1 (oC kg bnz/mol npht) = 2.77 oC;
tf = 5.5 oC – 2.77 oC = 2.7 oC.
0.326 mol Ca(NO3)2. x 