-a solution has a mixture of pentane and hexane ar roomtemp. The solution has a

Question

-a solution has a mixture of pentane and hexane ar roomtemp. The solution has a vapor pressure of 258 torr. Pure pentane and hexane have vapor pressures of 425 torr and 151torr, respectively at room temp. What is the mole fractioncomposition of the mixture? (assume ideal behavior) -the freezing point of t butanol is 25.50 deg C and Kf is 9.1deg C/m. Usually, t butanol asborbs water on exposure toair. If the freezing point of a 10.0 g sample of t-butanol is24.59deg  how many grams of water are present? -a solution has a mixture of pentane and hexane ar roomtemp. The solution has a vapor pressure of 258 torr. Pure pentane and hexane have vapor pressures of 425 torr and 151torr, respectively at room temp. What is the mole fractioncomposition of the mixture? (assume ideal behavior) -the freezing point of t butanol is 25.50 deg C and Kf is 9.1deg C/m. Usually, t butanol asborbs water on exposure toair. If the freezing point of a 10.0 g sample of t-butanol is24.59deg  how many grams of water are present?

Explanation / Answer

     Part A:          Thevapour pressure of solution = P = 258 torr                  Let the partial pressuresof pentane and hexane be p1 and p2               the vapour pressures of pure pentane and purehexane are 425 torr and 151 torr                      let p1o = 425 torr and p2o = 151 torr     let the mole fractions of pentane andhexane be X1 and X2                         p1 + p2 = p    --------------------   1                                  p1 + p2 = p    --------------------   1                        p1o X1 = p1  i.e   425 X1= p1   -------------   2                   p2oX2 = p2   i.e   151 X2 = p2  ------------ 3              We know that   X1 + X2 = 1             from equations 1 , 2 and 3           425 X1 + 151 X2 = p            425X1 + 151 ( 1 -X1 ) = 258       X1 = molefraction of pentane = 0.3905            X2 = mole fraction of hexane = 1 -0.3905 = 0.6095            Part B         Depression infreezing point = Tf = 25.50 - 24.59 =0.91oC              number of moles of t- Butanol = (10 / 74) = 0.1351          Tf = Kf x molality         0.91 = 9.1 x molality        molality =0.1       0.1 = (0.1351 x 1000)/ weight of solvent (g)      amount of water present = 1351g       

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