2) In lab 6, we measured the changes of gas volume vs. pressure. i) Using any of

Question

2) In lab 6, we measured the changes of gas volume vs. pressure. i) Using any of two sets of P-V data from your measurement to demonstrate that it agrees with the Boyle's law. 321004 031.2 ii) According to your experiment, what is the initial mole number (n) of air used in the entire measurement? ii From the initial state (when your started the measurement) to final state (when you ended the measurement), what is the system entropy change S for the process? 3) In exp. 7, we measured freezing point depression. i) Now if we dissolve 0.64 g solid naphthalene (CyoHs, MW- 128) in 10 g liquid cyclohexane (CH12, MW-84), what is the molality m of naphthalene? ii) Pure cyclohexane has freezing point 6.6 °C, and its freezing point depressing constant Kf-20.4°C/m. For the solution above, what is its freezing point theoretically? IF-Kfxm

Explanation / Answer

From the third experiment, where V= 8ml and P= 453.9, PV =453.9*8=3631.2 mm Hg.ml

From the 4th experiment, where V= 10ml, and P= 348.9, PV= 348.9*10=3489 mm Hg.ml

So PV is almost constant.

From gas law, PV= nRT, n = no of moles = PV/RT

P= pressure in atm from second data point, P= 543.4 mm Hg= 543.4/760 atm =0.715 atm, V= 6ml= 6/1000L= 6*10-3L, R= 0.0821 L.atm/mole.K, T= 25 deg.c, assumed= 25+273= 298K

n= 0.715*6*10-3/ (0.0821*298)=0.000175 moles

entropy change = -nR*ln(P2/P1), P2= pressrue in final stage= 175.6 and P1= initial pressure = 743.8

entropy change = -0.000175*8.314*ln(175.6/743.8)=0.0021 J/K

2. molality = moles of solute/1000 gm of solvent

moles of napthalene = mass of napthalene/ molar mass= .64/128= 0.005

mass of cyclohexane= 10 gm, 1000gm= 1kg, 10 gm= 10/1000 kg =0.01 kg

molality, m= 0.005/0.01 =0.5m

b) Freezing point depression = i*Kf*m, I =van't Hoff factor for cyclohexane= 1 ( non dissociating), Kf=20.4 deg.c/m

Freezing point depresion= 1*20.4*0.5= 10.2

Freezing point of the solution = freezing point of cyclohexane- freezing point depression =6.6-10.2=-3.6 deg.c

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