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`        The molar volume of liquid benzene increases by a factor of 2750 as it vaporizes at 20 C. At 27 C when a non volatile solute is dissolved in 54.6 cm3 of benzene vapour pressure of this solution is found to be 98.88mmHg. Calculate the freezing point of the solution Given Enthalpy of vapourisation of benzene = 394.57J/gEnthalpy of fusion of benzene(l) = 10.06KJ/molMolal depression constant for benzene= 5.0 K Kg/ Mol`
one year ago

Arun
23479 Points
```							Please find the answer to your question (i)Volume = No. of moles * molar mass/density(ii) PV = nRT or P = nRT/VVolume of 1 mole of liq. Benzene = 78/0.877Volume of 1 mole of toluene. 92/0.867In vapour phase,At 20oC, for 1 mole of benzene,Volume = 1 **78 *2750/0.877 = 244583.80 mL= 244.58 LSimilarly for 1 mole of toluene,Volume = 1 *92/0.867 * 7720= 819192.61 mL = 819.199 LAs we know that, PV = nrtFor benzene, PoB = nRT/V = 1 *0.0821 *293/244.58 atm= 0.098 atmFor toluene, PoT = nRT/V = 1 *0.0825 *293/819.19 atm = 0.029 atmP = PoB.XB + PoT.XT∵XB + XT = 1∴ XT= 1 - XBP = PoB.XB + PoT (1 – XB)Total vapour pressure = 46 torr = 46/760 = 0.060 atmThus, 0.060 = 0.098 XB + 0.029 (1 - XB)⇒ 0.060 = 0.098 XB + 0.029 – 0.029 XB⇒ 0.031 = 0.069 XB∴ XB = 0.031/0.069 = 0.45 (in liquid phase)XB + XT = 1XT = 1 – 0.45 = 0.55 (in liquid phase)Also, P’B = PoB.XB = PX’BSo, 0.098 * 0.045 = 0.060 * X’BX’B = 0.098 * 0.45/0.060 = 0.735 (in gas phase)
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one year ago
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