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Ethanol has a heat of vaporization of 38.56kJ/mol and a normal boiling point of 78.4 ℃

Ethanol has a heat of vaporization of 38.56kJ/mol and a normal boiling point of 78.4 ℃ .

What is the vapor pressure of ethanol at 17 ℃?

Express your answer using two significant figures. Units in torr

Solutions

Expert Solution

logP2/P1 = Hv/2.303R [1/T1 -1/T2]

T1            = 170C   = 17 + 273   = 290K

T2           = 78.40C    = 78.4 + 273 = 351.4K

P1          =

P2          = 760torr

Hv        = 38.56Kj/mole = 38560J/mole

logP2/P1 = Hv/2.303R [1/T1 -1/T2]

log760/P1 = 38560/2.303*8.314 [1/290 -1/351.4]

                  = 2013.87(0.003448-0.00284)

                  = 1.2244

log760/P1   = 1.2244

760/P1        = 10^1.2244

760/p1         = 16.7648

P1            = 760/16.7648 = 45.33 torr >>>>answer or 43 torr

queen_honey_blossom answered 2 years ago
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