CHEM 1411
PRACTICE PROBLEMS II (Chapters 4, 5)
Multiple Choices: Select one best answer.
Q1-11: Chapter 4; Q12-22: Chapter 5;
1. Which of the following is a weak electrolyte?
(a) barium hydroxide solution
(b) He gas
(c) ammonia solution
(d) potassium chloride
(e) hydrochloric acid
Hint: p.p. 120-121. Weak electrolytes include molecular compounds that are weak acids and weak
bases. Strong electrolytes include ionic compounds and molecular compounds that are strong acids.
Nonelectrolytes include all molecular compounds except strong acids, weak acids and weak bases.
2. Which of the following is insoluble in water?
(a) (NH4)2CO3
(b) NaI
(c) K2SO4
Hint: Refer to p.p.123, Table 4.2
(d) PbI2
(e) Ba(OH)2
3. Which of the following is wrong concerning a net-ionic equation?
(a) S2-(aq) + Cu2+(aq)  CuS(s)
(b) 2NaOH(aq) + MgSO4(aq) Mg(OH)2(s) + Na2SO4(aq)
(c) SO42-(aq) + Pb2+(aq) PbSO4(s)
(d) CO32-(aq) + Ca2+(aq) CaO + CO2(g)
2+
2+
(e) Fe(s) +Cu (aq)  Fe + Cu(s)
Hint: p.p. 124-126. Examples 4.1 & 4.2. Check with the solubility table for the products, which must
be solids (s), liquids (l) or gases (g).
4. Which of the following is a Brǿnsted acid?
(a) HCO3(b) CH3COO(c) S2(d) SO4(e) NO3Hint: p. 128. Example 4.3. Definition. Brǿnsted acids must contain available (releasable) hydrogen
ions while the Brǿnsted bases must be able to accept hydrogen ions.
5. Which of the following underlined atoms contains the oxidation number as -1?
(a) Cs2O
(b) CaC2
(c) SO42(d) PtCl42(e) Na2O
Hint: p.p. 134-137. Rules. Example 4.5.
6. Which of the following redox reactions will occur according to activity series?
(a) Cu(s) + 2HCl(aq)  CuCl2(aq) + H2(g)
(b) Zn(s) + CuSO4(aq)  ZnSO4(aq) + Cu(s)
(c) Fe(s) + ZnSO4(aq)  FeSO4(aq) + Zn(s)
(d) 2Al(s) + 3Ca(NO3)2(aq)  2Al((NO3)3(aq) + 3Ca(s)
Hint: Activity table, Fig. 4.6, p. 139. Any metal will be oxidized by the metal ion below it in the
activity series.
7. Which of the following is not the redox reaction?
(a) 2HCl  Cl2 + H2
(b) H2 + O2  H2O2
(c) CaCO3  CaO + CO2
(d) 2HNO3 + Mg  Mg(NO3) 2+ H2
(e) Cl2+NaBrNaCl+Br2
Hint: Redox reaction involves electron transfer, which leads to change of oxidation number of the
elements. You need first to figure out the oxidation number of each species. The reaction that does not
have change in oxidation number of elements is not a redox reaction.
8. How many grams of KOH are present in 35.0 mL of a 5.50 M solution?
(a) 5.5
(b) 10.8
(c) 15.7
(d) 17.8
(e) 21.3
Hint: p.p. 142-146. Example 4.6. In order to get grams, we need to get moles first. M=n/V,
n=MV=5.50M*(35.0/1000)L=0.193mole, m=nM=0.193mole*56.1g/mole=10.8g
9. What is the molarity (M) of a solution made by 5.85 g NaCl in 200.0 mL of solution?
(a) 0.4
(b) 0.5
(c) 0.8
(d) 0.9
(e) 1.4
Hint:molarity=n/V, n=5.85g/(58.44g/mole)=0.1mole, molarity= 0.1mole/(200/1000)L=0.5M
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10. What is the final concentration (in M) of a solution when water is added to 25.0 mL of a 0.866 M
KNO3 solution until the volume of the solution is exactly 500.0 mL?
(a) 0.0252
(b) 0.0368
(c) 0.0117
(d) 0.0534
(e) 0.0433
Hint: p.p. 146-148. Similar to the reverse question of Example 4.8. MiVi=MfVf,
Mf=MiVi/Vf= 25.0ml*0.866M/500.0ml=0.0433M
11. What volume (in mL) of a 0.500 M HCl solution is needed to neutralize 10.0 mL of a 0.2000 M
Ba(OH)2 solution?
(a) 8.00
(b) 4.00
(c) 2.00
(d) 1.00
(e) 0.50
Hint: p.p. 150-153. Solution stoichioimetry. Example 4.11. Short-cut formula of acid-base
neutralization: ia x Ma x Va = ib x Mb x Vb where a refers to acid and b refers to base. ia refers to number
of H in the chemical formula of acid and ib refers to the number of OH in the chemical formula of base.
12. What volume (in L) does a sample of air occupy at 6.6 atm when 1.2 atm, 3.8 L of air is
compressed?
(a) 0.34
(b) 0.57
(c) 0.69
(d) 0.77
(e) 0.86
Hint: Boyle’s law. p.p. 175-178. Equation (5.2). PV= constant. P1V1=P2V2,
V2= P1V1/P2=1.2atm*3.8L/6.6atm = 0.69L
13. What is the final temperature (in K), under constant-pressure condition, when a sample of
hydrogen gas initially at 88oC and 9.6 L is cooled until its finial volume is 3.4 L?
(a) 31
(b) 68
(c) 94
(d) 128
(e) 260
Hint: Charles’s Law. p.p. 178-180. Equation (5.4). V/T= constant, V1/T1=V2/T2,
T2=V2T1/V1= 3.4L*(88+273.15)K/9.6L=128K
14. What is the volume (in L) of 88.0 g of carbon dioxide gas at STP?
(a) 44.8
(b) 53.7
(c) 62.1
(d) 74.6
(e) 83.2
Hint: Ideal gas law. p.p. 181-182. Equation (5.8). Example 5.3. T=0◦C, P=1atm,
n=m/M=88.0g/(44g/mole)= 2.0 mole, PV=nRT, V=nRT/P= 2.0mole*0.082 L• atm / (mol •
K)*273.15K/1atm=44.8L
15. What is the pressure of the gas (in atm) when 5.0 moles of CO gas are present in a container of
20.0 L at 27 oC?
(a) 3.25
(b) 6.15
(c) 7.40
(d) 9.30
(e) 10.55
Hint: PV=nRT, P=nRT/V= 5.0mole*0.082 L• atm / (mol • K)*(27+273.15)K/20.0L=6.15atm
16. A gas evolved during the fermentation of glucose (wine making) has a volume of 0.78 L at 20.1 oC
and 1.00 atm. What was the volume (L) of this gas at the fermentation temperature of 36.5 oC and 2.00
atm pressure?
(a) 0.41
(b) 0.82
(c) 1.43
(d) 2.67
(e) 3.54
Hint: Combined gas law. p. 184 equation (5.10). PV/T=Constant, P1V1/T1=P2V2/T2;
V2=P1V1T2/T1P2=1.00atm*0.78L*(36.5+273.15)K/[(20.1+273.15)K*2.00atm]=0.41L
17. What is the molar mass (g/mol) of 7.10 grams of gas whose volume is 5.40 L at 741 torr and 40 oC?
(a) 34.6
(b) 70.3
(c) 86.2
(d) 94.6
(e) 102.3
Hint: p. 187. Equation (5.12). Example 5.9. M=dRT/P=mRT/VP=7.10g*0.082 L• atm / (mol • K)
*(40+273.15)K /[(741/760)atm*5.40L]= 34.6
18. What is the density of HCl gas in grams per liter at 700 mmHg and 25 oC?
(a) 0.54
(b) 1.37
(c) 2.24
(d) 2.97
(e) 3.57
Hint:p.p.186, d=m/V=PM/RT=(700/760)atm*36.45g/mole/[0.082 L• atm / (mol • K)*(25+273.15)K]
=1.37g/L
19. A compound has the empirical formula as SF4. At 20 oC, 0.100 gram of the gaseous compound
occupies a volume of 22.0 mL and exerts a pressure of 1.02 atm. What is the molecular formula of the
gas?
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(a) SF4
(b) SF6
(c) S2F10
(d) S4F16
(e) S5F20
Hint: p. 211 (5.50) or p. 187. Equation (5.12). Example 5.9. In order to get the molecular formula, we
need to get the molar mass of the compound first. M=dRT/P=mRT/VP
=0.100g*0.082 L• atm / (mol • K)*(20+273.15)K/[(22.0/1000)L*1.02atm]=107.1g/mole. The molar
mass of the empirical formula SF4 is 32.07+4*19.00=108.07g/mole. This is very similar to the
calculated value 107.1g/mole. So the molecular formula is SF4.
20. The combustion process for methane, major component of natural gas, is
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(l)
If 15.0 moles of methane are reacted, what is the volume of carbon dioxide (in L) produced at 23.0 oC
and 0.985 atm?
(a) 369.8
(b) 430.7
(c) 510.8
(d) 630
(e) 720
Hint: Gas stoichiometry: p.p. 190-192. Example 5.11. This question is taken from p. 211 (5.52).
x 1CO2/1CH4
15 mole CH4 ---------------- 15 mole CO2
PV=nRT, V=nRT/P=15mole*0.082 L• atm / (mol • K)*(23.0+273.15)K/0.985atm=369.8L
20. In alcohol fermentation, yeast converts glucose to ethanol and carbon dioxide:
C6H12O6(s)  2 C2H5OH(l)) + 2 CO2(g)
If 5.97 g of glucose are reacted and 1.44 L of carbon dioxide gas are collected at 293 K and 0.984 atm,
what is the percent yield of the reaction?
(a) 89.4 %
(b) 76.3%
(c) 65.9%
(d) 56.2%
(e) 47.6%
Hint: Gas stoichiometry: p.p. 190-192. This question is taken from p. 212 (5.54).%Y=Real
yield/theoretical yield; theoretical yield can be obtained from reaction of glucose
/180g/mole
*2CO2/1 C6H12O6
C6H12O6 5.97g ------------------0.033mole----------------------0.066mole-----------
Real yield can be obtained though the collection of CO2,PV=nRT,
n = PV/RT= 0.984atm*1.44L/[0.082 L• atm / (mol • K)*293K]=0.059mole;
%Y= 0.059/0.066=0.894 =89.4%
21. A mixture of gases contains 0.31 mol CH4, 0.25mol C2H6 and 0.29 mol C3H8. The total pressure is
1.50atm. Calculate the partial pressure of the CH4, C2H6 and C3H8 gases
(a) 0.23,0.35,0.56atm
(b) 0.54,0.44,0.51atm
(c) 0.54,0.44,0.51atm
(d) 0.54,0.44,0.51atm
(e) 0.54,0.44,0.51atm
Hint : group practice problem Chap5, Q5.63 end of chapter problems. P i=PTXi,
nT= 0.31+0.25+0.29=0.85mole; Xi=ni/nT;
P CH4= 1.50atm*(0.31mole/0.85mole)=0.54atm; P C2H6 = 1.50atm*(0.2 mole/0.85mole)=0.44atm;
P C3H8= 1.50atm*(0.29mole/0.85mole)=0.51atm;
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