PART ll. Longer Questions (46%) 1410 points) Pure benzophenone freezes at 48.1 .
ID: 695460 • Letter: P
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PART ll. Longer Questions (46%) 1410 points) Pure benzophenone freezes at 48.1 . A solution of 1.25 g of biphenyl (Cillo; molar mass - 154.2 g/mol) with 25.38 g of benzophenone froze at 16.8 °C. Calculate the molality of the solution (m) and the molal freezing point depression constant (r). Show all calculations 2. (10 points). When zine chloride (ZnCl) reacts with NHs solution, a white crystalline compound Zn(NHsCla is produced. A student used the procedure in this module to determine the mass percent of NH, in Zn(NH ).Cl. The student added 25.0 ml of 0.2500 M HCI solution to an aqueous solution containing 0.3145 g of Zn(NH) Clh. The excess HCI required 25.60 ml of 0.1000 M NaOH solution for complete titration. Caleulate the mass percent of NH, in Zn(NHs) Cl based on the student's data 3. (10 points) Pure camphor freezes at 178.4 °C and its molal freezing point depression constant (Ki) is 37.7°C kg/mol. A solution containing 2.1 g of unknown and 12.46 g of camphor froze at 168.4 °C. Calculate the molar mass of the unknown and show all calculations.Explanation / Answer
1.
Mass of biphenyl = 1.25 g.
Molar mass of biphenyl = 154.2 g/mol
Amount of bihenyl = mass / molar mass = 1.25 / 154.2 = 0.00811 mol
Mass of benzophenone = 25.38 g. = 0.02538 kg.
Molality of solution = moles of solute / mass of solvent in kg
m = 0.00811 / 0.02538
m = 0.3194 m
Depression in freezing point = Kf * m
48.1 - 16.8 = Kf * 0.3194
Kf = molar freezing point depression constant = 98.0 kg.mol-1.0C-1
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