In 100 g of pure benzene whose freezing and boiling temperatures are 5.5 100C and 80.1 C respectively, 24.4 g biphenyl (C12H10) is dissolved. Since the cryoscopy and ebulioscopy constants of benzene are 5.12 K.kg/mol and 2.63 K.kg/mol respectively; Find the vapor pressure of the solution at 80.1ᴼC, the freezing and boiling temperatures of benzene in the solution. (MAbenzene = 78 g / mol; MAbiphenyl = 154 g / mol; 1atm = 760 mmHg)
In 100 g of pure benzene whose freezing and boiling temperatures are 5.5 100C and 80.1 C respectively, 24.4 g biphenyl (C12H10) is dissolved. Since the cryoscopy and ebulioscopy constants of benzene are 5.12 K.kg/mol and 2.63 K.kg/mol respectively; Find the vapor pressure of the solution at 80.1ᴼC, the freezing and boiling temperatures of benzene in the solution. (MAbenzene = 78 g / mol; MAbiphenyl = 154 g / mol; 1atm = 760 mmHg)
Principles of Modern Chemistry
8th Edition
ISBN:9781305079113
Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Publisher:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Chapter11: Solutions
Section: Chapter Questions
Problem 47P
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In 100 g of pure benzene whose freezing and boiling temperatures are 5.5 100C and 80.1 C respectively, 24.4 g biphenyl (C12H10) is dissolved. Since the cryoscopy and ebulioscopy constants of benzene are 5.12 K.kg/mol and 2.63 K.kg/mol respectively; Find the vapor pressure of the solution at 80.1ᴼC, the freezing and boiling temperatures of benzene in the solution. (MAbenzene = 78 g / mol; MAbiphenyl = 154 g / mol; 1atm = 760 mmHg)
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