A solution is made by mixing 33.0 ml of ethanol, C2H6O and 67.0 ml of water. Assuming ideal behavior, what is the vapor pressure
of the solution at 20 degree C? Ethanol= .789 g/ml 43.9 torr. water= .998 g/ml 17.5 torr
1 answer:
Grams ethanol = 33 ml times .789 gms/ml = 26.037 gms
<span>Moles ethanol = 26.037 gms / 46 gms/mole = .57 moles </span>
<span>Moles water = 67 ml or 67 grams/18 gms/mole = 3.22 moles </span>
<span>total moles = .57 + 3.72 = 4.29 moles </span>
<span>Mole fraction ethanol = .57 moles ethanol / 4.29 moles total = 0.13</span>
<span>Moles fraction water = 3.72 moles water / 4.29 moles total = 0.87</span>
<span>Partial pressure of ethanol = mole fraction ethanol (.13) _ times VP ethanol 43.9 torr) = 5.707 torr </span>
<span>partial pressure water = mole fraction water .87) times VP water (l7.5 torr) = 15.23 torr </span>
<span>Total vapor pressure over solution = 5.71 torr + 15.23 torr = 20.94 torr</span>
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