At a giventemperature, you have a mixture of benzene (vapor pressure of purebenz

Question

At a giventemperature, you have a mixture of benzene (vapor pressure of purebenzene = 745 torr) and toluene (vapor pressure of pure toluene =290 torr). The mole fraction of benzene in the solution is0.590. Assuming ideal behavior, calculate the mole fractionof toluene in the vapor above the solution. 0.213 0.778 0.641 0.359 0.590 At a giventemperature, you have a mixture of benzene (vapor pressure of purebenzene = 745 torr) and toluene (vapor pressure of pure toluene =290 torr). The mole fraction of benzene in the solution is0.590. Assuming ideal behavior, calculate the mole fractionof toluene in the vapor above the solution. 0.213 0.778 0.641 0.359 0.590

Explanation / Answer

Given that Vapor pressure of pure benzene = 745 torr    Mole fraction benzene = 0.59 Mole fraction of toluene = 0.41 Vapor pressure of pure toluene = 290 torr The partial pressure of benzene=    Mole fraction benzene *Vapor pressure of purebenzene                                                        =0.59 * 745 torr                                                       = 439.55 torr        The partial pressureof  toluene =    Molefraction  toluene *Vapor pressure ofpure  toluene                                                        =0.41 * 290 torr                                                       = 118.9 torr    The molefractionof  toluene in vapor = The partial pressureof  toluene/ (The partial pressureof  toluene +The partial pressure of benzene )                                                               = 118.9 torr / (118.9 torr + 439.55 torr)                                                               = 0.2129 Part II Moles of NaCl = mass / molar mass                          = 150.0 g / 58.453 g/mol                           =2.566 mol It is dissociated completely so total moles of soluteparticles in solution = 2* 2.566 mol                                                                                                             = 5.1323 mol    Moles of water = 1000g/18g/mol                            =55.56 mol    Moles fraction of water = moles of water / totalmoles                                        = 55.56 mol / (55.56mol + 5.1323mol)                                        = 0.9154    The vapor pressure of solution =mole fraction of water * vapor pressure of pure water                                                       = 0.9154 * 23.8 torr                                                       = 21.78 torr Part II Moles of NaCl = mass / molar mass                          = 150.0 g / 58.453 g/mol                           =2.566 mol It is dissociated completely so total moles of soluteparticles in solution = 2* 2.566 mol                                                                                                             = 5.1323 mol    Moles of water = 1000g/18g/mol                            =55.56 mol    Moles fraction of water = moles of water / totalmoles                                        = 55.56 mol / (55.56mol + 5.1323mol)                                        = 0.9154    The vapor pressure of solution =mole fraction of water * vapor pressure of pure water                                                       = 0.9154 * 23.8 torr                                                       = 21.78 torr

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