10 CM.014 My A sample of 58.50 g solid benzene at -2.4°C was heated until it a f

Question

10 CM.014 My A sample of 58.50 g solid benzene at -2.4°C was heated until it a final phase and final temperature of the benzene sample bsorbed 13.5 kJ of heat t, at a constant pressure of 1 atmosphere. Using the data given in the table below, determine the Values 5.49°C 80.090 9.87 k)/mol 20.72 k/mol melting point boiling point at 5.49C heat of vaporization at s0.0g°C at 0°C 118.4 /mol K molar heat capacity of liquid at 25°C 136.0 J/mol K molar heat capacity of gas at 95 C 103.98 1/mol-K will be oc Supporting Materials Type here to 906 PM 2/18/2018

Explanation / Answer

Benzene at -2.4 oC would be a solid

heat absorbed (q) = 13.5 kJ = 13500 J

mass benzene = 58.50 g

initial temperature = -2.4 oC

let Tf be the final temperature

Step 1 : benzene solid (at -2.4 oC) to solid (at 5.49 oC),

q1 = mCpdT = 58.50 x (118.4/78) x (5.49 + 2.4) = 700.632 J

Step 2 : benzene solid (at 5.49 oC) to liquid (at 5.49 oC)

q2 = mdHf = 58.50 x (9870/78) = 7402.5 J

Total heat remains now = 13500 - (q1 + q2) = 5397 J

Now,

step 3 : liquid (at 5.49 oC) to liquid (at Tf oC)

5397 = 58.50 x 136/78 x (Tf - 5.49)

5397 = 102Tf - 560

Tf = 5957/102 = 58.4 oC

Thus, the final temperature of benzene is 58.4 oC and is in liquid phase.

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