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Answers to Chapter 05 Study Questions

    1. STP:  \(\mathrm{P_T = 1\: atm}\); \(\mathrm{T = 273\: K}\);  \(\mathrm{P_{O_2}=\left(\dfrac{n_{O_2}}{n_T}\right)P_T}\);  \(\mathrm{n_T = 0.039 + 0.010 + 0.001 = 0.050\: moles}\)
      \(\mathrm{P_{O_2}=\left(\dfrac{n_{O_2}}{n_T}\right)P_T}\); \(\mathrm{P_{O_2}=\left(\dfrac{0.010}{0.050}\right)(1.00\:atm)=0.20\: atm}\)
    2. \(\mathrm{V  =  ?}\);  STP:  \(\mathrm{T = 273\: K}\), \(\mathrm{P_T = 1\: atm}\);  \(\mathrm{n_T = 0.050\: mol}\);  \(\mathrm{PV = nRT}\)
      \(\mathrm{V=\dfrac{nRT}{P}=\dfrac{(0.050\:mol)(0.08206)(273\:K)}{1\:atm}=1.1\:L}\)

 

    1. \(\mathrm{P_T  = P_{H_2} + P_{H_2O}}\); Find \(\mathrm{P_{H_2O}}\) in Table from lab report; at 19°C, \(\mathrm{P_{H_2O} = 16\: mmHg}\)
      \(\mathrm{P_{H_2} =  P_T  -  P_{H_2O}  =  756 - 16  =  740.\: mmHg}\)
    2. \(\mathrm{740\: mmHg\times\dfrac{1\:atm}{760\:mmHg}=0.974\: atm}\)

 

  1. \(\mathrm{V_1 = 600.\: cm^3}\);  \(\mathrm{T_1 = 25^\circ C = 298\: K}\);  \(\mathrm{P_1 = 750.\: mmHg}\)

\(\mathrm{V_2 = 480.\: cm^3}\);  \(\mathrm{T_2 = 41^\circ C = 314\: K}\);  \(\mathrm{P_2 = ?}\)

\(\mathrm{P_2=P_1\times\dfrac{V_1}{V_2}\times\dfrac{T_2}{T_1}=750\:mmHg\times\dfrac{600\:cm^3}{480\:cm^3}\times\dfrac{314\:K}{298\:K}=988\: mmHg}\)

 

  1. \(\mathrm{density =\dfrac{molar\:mass}{molar\:volume}=\dfrac{4.00\:g}{22.4\:L}=0.178\: g/L}\)

 

    1. \(\mathrm{2C_4H_{10}(g) + 13O_2(g) \rightarrow 10H_2O(l) + 8CO_2(g)}\)
    2. \(\mathrm{2.0\: L\: CO_2\times\dfrac{13\:L\:O_2}{8\:L\:CO_2}=3.2\: L\: O_2}\)
    3. \(\mathrm{11.6\: g\: C_4H_{10}\times\dfrac{1\:mol\:C_4H_{10}}{58.0\:g\:C_4H_{10}}\times\dfrac{8\:mol\:CO_2}{2\:mol\:C_4H_{10}}\times\dfrac{22.4\:L\:CO_2}{1\:mol\:CO_2}=17.9\: L\: CO_2}\)
    4. \(\mathrm{5.6\: L\: C_4H_{10}\times\dfrac{1\:mol\:C_4H_{10}}{22.4\:L}\times\dfrac{10\:mol\:H_2O}{2\:mol\:C_4H_{10}}\times\dfrac{6.02\times10^{23}\:molecules}{1\:mol\:H_2O}=7.5\times10^{23}\:molecules}\)

 

  1. \(\mathrm{n = 1\: mole}\);  \(\mathrm{T = 68^\circ C = 341\: K}\);  \(\mathrm{P = 2.00\: atm}\);  \(\mathrm{V = ?}\)

\(\mathrm{V=\dfrac{nRT}{P}=\dfrac{(1\:mol)(0.08206)(341\:K)}{2.00\:atm}=14.0\: L}\)

 

  1. \(\mathrm{8.00\:g\:CH_4\times\dfrac{1\:mol\:CH_4}{16.0\:g\:CH_4}\times\dfrac{22.4\:L}{1\:mol}=11.2\: L}\)

 

  1. \(\mathrm{d=\dfrac{mm}{mV}}\); \(\mathrm{mV=\dfrac{nRT}{P}=\dfrac{(1\:mol)(0.08206)(546\:K)}{4.00\:atm}=11.2\:L}\);  \(\mathrm{d=\dfrac{44.0\:g}{11.2\:L}=3.93\: g/L}\)

 

  1. Find molar volume at 710 mmHg and 36°C and then use conversion factors:

T = 36 + 273 = 309 K;  \(\mathrm{P = 710\: mmHg\times\dfrac{1\:atm}{760\:mmHg}=0.934\:atm}\)

\(\mathrm{V=\dfrac{nRT}{P}=\dfrac{(1\:mol)(0.08206)(309\:K)}{0.934\:atm}=27.1\: L}\)

\(\mathrm{6.52\: g\: Cu S\times\dfrac{1\:mol\:CuS}{95.6\:g\:CuS}\times\dfrac{2\:mol\:O_2}{1\:mol\:CuS}\times\dfrac{27.1\:L\:O_2}{1\:mol\:O_2}=3.70\: L\: O_2}\)

 

  1. \(\mathrm{molar\: mass =\dfrac{mass}{moles}}\);  so use \(\mathrm{PV = nRT}\) to calculate \(\mathrm{n}\);  \(\mathrm{T = 29 + 273 = 302\: K}\);  \(\mathrm{P = 1\: atm}\)

\(\mathrm{n=\dfrac{PV}{RT}=\dfrac{(1\:atm)(6.20\:L)}{(0.08206)(302\:K)}=0.250\:moles}\);  \(\mathrm{molar\:mass=\dfrac{7.00\:g}{0.250\:mol}=28.0\: g/mole}\)

 

  1. \(\mathrm{15.0\: g\: CO_2\times\dfrac{1\:mol\:CO_2}{44.0\:g\:CO_2}=0.341\: mol\: CO_2}\);   \(\mathrm{12.0\:g\:CH_4\times\dfrac{1\:mol\:CH_4}{16.0\:g\:CH_4}=0.750\: mole\: CH_4}\); 

At constant T and P, \(\mathrm{\dfrac{V_1}{n_1}=\dfrac{V_2}{n_2}}\);   \(\mathrm{V_2=V_1\times\dfrac{n_2}{n_1}=7.16\:L\times\dfrac{0.750\:mol}{0.341\:mol}=15.7\:L}\)

 

  1. \(\mathrm{P_{H_2}= P_T - P_{H_2O}}\); At 22°C, \(\mathrm{P_{H_2O} = 20\: mm Hg}\); \(\mathrm{P_{H_2} = 750\: mmHg - 20\: mmHg = 730\: mmHg}\)

\(\mathrm{P_{H_2}  = 730\: mmHg \times\dfrac{1\:atm}{760\:mmHg} = 0.960\: atm}\);  T = 22 + 273 = 295 K;

\(\mathrm{7.78\: g\: Zn \times\dfrac{1\:mol\:Zn}{65.4\:g\:Zn}\times\dfrac{1\:mol\:H_2}{1\:mol\:Zn}=0.119\: mol\: H_2 =  n}\)

\(\mathrm{V=\dfrac{nRT}{P}=\dfrac{(0.119\:mol)(0.08206)(295\:K)}{0.960\:atm}=3.00\: L}\)