Solutions 61. The volume of the laboratory isV = 5.0×5.0×3.0 = 75.0 m3a) water: p = 24 Torrn = pV/RT = (24 Torr) × (75.0×103 L) / {(62.364 L Torr K-1 mol-1) ×(298.15 K)} = 96.855 molm = nM = (96.855 mol) × (18.02 g mol-1) = 1745.3 g = 1.7453 kgb) benzene: p = 98 Torrn = pV/RT = (98 Torr) × (75.0×103 L) / {(62.364 L Torr K-1 mol-1) ×(298.15 K)} = 395.49 molm = nM = (395.49 mol) × (78.11 g mol-1) = 30890.8 g = 30.89 kgc) mercury: p = 1.7×10-3 Torrn = pV/RT = (1.7×10-3) × (75.0×103 L) / {(62.364 L Torr K-1 mol-1) ×(298.15 K)} = 0.00686 molm = nM = (0.00686 mol) × (200.59 g mol-1) = 1.376 g2. a) p1 = 10 Torr, T1 = 358.95 K and p2 = 40 Torr, T2 = 392.45 KUse the Clausius-Clapeyron equation to calculate ΔvapH:€ lnp2p1=ΔvapHR1T1−1T2      € ΔvapH =R ln p2p1( )1 T1−1 T2( )ΔvapH = (8.3145 J K-1 mol-1) × ln(40/10) / {1/(358.95 K) – 1/(392.45 K)}= 48.469 kJ mol-1 b) p3 = 760 Torr T3 - ?€ lnp3p1=ΔvapHR1T3−1T2      € T3= 1 1 T1−R ln p3p1( )ΔvapH      T3 = 1/{1/(358.95 K) - (8.3145 J K-1 mol-1) × ln(760/10) / (48.469×103 Jmol-1)} = 489.47 K = 216.32°C c) ΔvapS = ΔvapH/Tb = (48.469×103 J mol-1)/(489.47 K) = 99.0 J K-1 mol-13. ΔfusH = 2.292 kJ mol-1, T* = 234.3 K, ΔfusV = +0.517 cm3 mol-1, p* = 101.325 kPa.The additional pressure on the bottom of mercury column:Δp = p – p* = ρgh = (13.6×103 kg m3) × (9.81 m s-2) × (10 m) =1334160 Pahttps://www.coursehero.com/file/11467740/Homework-6-Solution-on-Physical-Chemistry/This study resource wasshared via CourseHero.com€ p = p*+ΔfusHΔfusVlnTT*€ lnTT*=p − p*( )ΔfusVΔfusHln(T/T*) = (1334160 Pa) × (0.517×10-6 m3 mol-1) / (2.292×103 J mol-1) =3.0×10-4T = (234.3 K) × exp(3.0×10-4) = 234.37 K4. We assume that a liter of seawater contains roughly 1000 g of water. Thennwater = 1000 g / 18.02 g mol-1 = 55.5 molnsolutes = 2×0.50 mol = 1.00 molxwater = nwater / (nwater + nsolutes) = 55.5/56.5 = 0.982pwater = xwater×pwater* = 0.982×2.338 kPa = 2.296 kPa5. Check whether pB/xB is equal (at least, approximately) to a constant (KB)x 0.005 0.012 0.019p/x 6.4×1036.408×1036.410×103Hence, KB ≅ 6.4×103 kPa6. We plot h/c against c:c3.2214.6185.1126.722h5.7468.2389.11911.99h/c1.7839181.7838891.7838421.783695y = -7E-05x + 1.7842R2 = 0.90391.783651.78371.783751.78381.783851.78391.783951.7840 2 4 6 8The intercept is 1.7842×10-2 m4 kg-1Intercept = RT/ρgM M = RT/(ρg×Intercept)M = (8.31451 J K-1 mol-1 × 293.15 K) / (1000 kg m-3 × 9.807 m s-2 × 1.7842×10-2m4 kg-1) = 13.93 kg mol-1 = 13.93 kDa7. Ebullioscopic constant:€ ΔT = KxB€ K =RT*2ΔvapHhttps://www.coursehero.com/file/11467740/Homework-6-Solution-on-Physical-Chemistry/This study resource wasshared via CourseHero.com€ ΔT = Kbb€ xB=nBnB+ nCCl4≈nBnCCl4=nBmCCl4mCCl4nCCl4= bMCCl4Therefore,€ Kb= KMCCl4=RT*2MCCl4ΔvapHKb = (8.3145 J K-1 mol-1)×(349.9 K)2×(153.81×10-3 kg mol-1)/(30.0×103 J mol-1) =5.22 K kg mol-1Cryoscopic constant:€ Kf=RT*2MCCl4ΔfusHKf = (8.3145 J K-1 mol-1)×(250.3 K)2×(153.81×10-3 kg mol-1)/(2.47×103 J mol-1) =32.4 K kg mol-18. According to Raoult’s law,pA = xA pA*Therefore, we can calculate mole fraction of benzene:xA = pA / pA* = 386 Torr / 400 Torr = 0.965€ xA=nAnA+ nB=mAMAmAMA+ mBMB=mAmA+ MAmBMB€ MB=xAMA1− xA( )mAmBMB = 0.965 × (78.11 g mol-1) × (19 g) / {(1 – 0.965) × (500 g)} = 82 g mol-19. y(N2) = 0.78 p = 760 TorrAccording to Dalton’s law, the partial pressure of N2 is calculated asp(N2) = 0.78×760 Torr = 592.8 TorrHenry’s law: p(N2) = K(N2)x(N2) x(N2) = p(N2)/K(N2)x(N2) = 592.8 Torr / 6.51×107 Torr = 9.1×10-6b(N2) = x(N2)/M(H2O) – see problem 4 for the relation between mole fraction andmolality.b(N2) = 9.1×10-6 / (18.02×10-3 kg mol-1) = 5.05×10-4 mol kg-1 = 0.51 mmol kg-1y(O2) = 0.21 p = 760 Torrp(O2) = 0.21×760 Torr = 159.6 Torrp(O2) = K(O2)x(O2) x(O2) = p(O2)/K(O2)x(O2) = 159.6 Torr / 3.30×107 Torr = 4.8×10-6b(O2) = x(O2)/M(H2O).b(O2) = 4.8×10-6 / (18.02×10-3 kg mol-1) = 2.68×10-4 mol kg-1 = 0.27 mmol kg-110. Kf(water) = 1.86 K kg mol-1https://www.coursehero.com/file/11467740/Homework-6-Solution-on-Physical-Chemistry/This study resource wasshared via CourseHero.comIn 250 cm3 of water approximately 250 g: m(water) = 250 gM(sucrose) = M(C12H22O11) = 342.3 g mol-1n(sucrose) = m(sucrose)/M(sucrose) = 7.5 g / 342.3 g mol-1 = 0.0219 molb = n(sucrose) / m(water) = 0.0219 mol / 0.25 kg = 0.0876 mol kg-1ΔT = Kfb = (1.86 K kg mol-1) × (0.0876 mol kg-1) = 0.16 KThe freezing temperature decreases by 0.16 K or 0.16°C.https://www.coursehero.com/file/11467740/Homework-6-Solution-on-Physical-Chemistry/This study resource wasshared via CourseHero.comPowered by TCPDF