Question 21 A A mixture contains 25 g of cyclohexane (CgH12) and 44 g of 2-methy

Question

Question 21 A A mixture contains 25 g of cyclohexane (CgH12) and 44 g of 2-methylpentane (C6H4). The mixture of liquids is at 35 °C. At this temperature, the vapor pressure of pure cyclohexane is 150 torr, and that of pure 2-methylpentane is 313 torr. Assume this is an ideal solution. What is the mole fraction of cyclohexane in the liquid phase? Incorrect Mark 0.00 out of 2.00 Flag question Answer: 3 Question 22 ncorrect Mark 0.00 out of 2.00 What is the vapor pressure (in torr) of cyclohexane above the solution? Answer: 51 Flag question Question 23 ncorrect Mark 0.00 out of 2.00 What is the mole fraction of cyclohexane in the vapor phase? Answer: 2 Flag question

Explanation / Answer

moles of cyclohexane = 25/ M.W = 0.297

moles of methyl pentane = 44 / M.W = 0.51

mole fraction of cyclohexane in liquid phase = 0.368

mole fraction of methylpentane in liquid phase = 0.632

The vapour pressure of a solution of a non-volatile solute is equal to the vapour pressure of the pure solvent at that temperature multiplied by its mole fraction.

p for cyclohexane = Po * x

p = 150 * 0.368 = 55.2 torr

p for methyl pentane = Po * x

p = 313* 0.632 = 197.816 torr

Since PV = nRT, then moles of vapor are proportional to vapor pressure, and so the mole fraction of a component is equal to the "pressure fraction" of the vapor pressures:
55.2 / (55.2 + 197.816) = 0.218 mole fraction of cyclohexane

osmotic pressure = iMRT

= 2 * 5.3*0.0821 *305 / 58.44 = 4.54 atm

delta T = i*Kf*m

delta T = 1*4.9*1.45 = 7.105

freezing point =5.51 - 7.105 = -1.595 C

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