Pre-Laboratory Assignment ed MISC 327, Graphical Representation of Data, in this

Question

Pre-Laboratory Assignment ed MISC 327, Graphical Representation of Data, in this describes the principles of graphing. Representation of Data, in this series or another authoritative source that 2 A student beginning this experiment accidentall laboratory bench. Describe any potentia method of cleaning up from the accident. ent accidentally spilled some t-butyl alcohol on his hands and on the student begmn Describe any potential danger this situation might cause and state the proper of nitrobenzene and a nonionic unknown was used to molar mass of the unknown. Time-temperature data for the cooling of nitrobenzene 3. The freezing point depression of a solution determine the and for the cooling of a solution containing 50.0 g of nitrobenzene and 5.00 mL of a nonionic liquid unknown, are given belovw 6.87 "C Kg mol- . The density of the unknown was 0.714 g mL . The Ky of nitrobenzene is nitrobenzene +unknown time, min temp, °C 12.50 10.75 time, min temp, C 9.00 1.5 1.5 7.00 4.80 4.50 3.0 4.0 5.0 4.50 4.0 1.25 5.00 5.30 5.30 5.20 -2.25 -2.50 7.0 7.0 9.0 10.0 11.0 12.0 9.0 10.0 11.0 12.0 13.0 14.0 15.0 16.0 1.50 14.0 15.0 16.0 -2.00 -2.10

Explanation / Answer

2) The freezing point of Nitrobenzene is the lowest point after which the slope changes so it is 5.30 oC.

4) The freezing point of the mixture of nitrobenzene + unknown is the lowest point after which the slope changes so it is -2.50.

5) Freezing point depression is 5.30 - (-2.50) = 7.80oC

6) molality of the solution is defined as moles per 1 Kg of the solvent.

depression in freezing point = Kf x molality

7.80 = 6.87 x molality

molality = 7.80/6.87 = 1.13

6) molality = moles/Kg of solvent

5mL x density = 5 mL x 0.714 = 3.57 g in 50 g of nitrobenzene

so in 1 Kg we will have 1000 x 3.57/50 = 71.4 g

71.4 g will be 1.13 moles since molality is 1.13

so molar mass will be 63.2 g/mol

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