I had to perform an experiment for the freezing point depression in which we had to determine the freezing points of water as a pure solvent, sodium chloride, sucrose, and ethylene glycol.
We had to work with a partner so I completed the experiment for water, NaCl, and the glycol and had to get the info for sucrose from my partner. The data that was collected was the freezing pint of DI water which was the corrected value for the thermometer that I used. We had to record the temp every 30 seconds until we saw ice crystals. We had to do the same thing for the NaCl, sucrose, and gycol. For the NaCl we started at 0.5 min in which the temp was 15.0 degrees celsius and crystals formed at 10.5 min at -2.0 degrees celsius.
We had to measure out 5.800g of NaCl and record the exact weight. First I weighed the measuring boat and then I poured in the NaCl so to get the exact weight I subtracted the mass of the measuring boat with NaCl from the mass of just the boat. Is this correct?
Also, I need to calculate the # of moles of solute for NaCl and sucrose. To do this for NaCl would I divide the mass of solute by the molecular weight of NaCl?
A freezing point depression experiment was conducted using cyclohexane as the solvent. The freezing point of pure cyclohexane is 6.60°C and the freezing point depression constant is 20.00°C/m. The freezing point of a solution
The freezing-point depression for a given aqueous solution is 0.32 K. The freezing-point depression constant for water is 1.86 K/m. Calculate the molality of solutes in the solution. The equation for freezing point depression is
1. Find the molality of the solution prepared by dissolving 0.238g toluene, C7H8, in 15.8g cyclohexane 2. A pure sample of the solvent phenol has a freezing point of 40.85 degrees C. A 0.414 molal solution of isopropyl alcohol was
The procedure described in this experiment was used to determine the molar mass of unknown liquid (non-electrolyte). The solution was made by mixing 0.961 g of the unknown with 100.0 g of water. The freezing point depression of
The freezing point of a 0.010 m aqueous solution of a nonvolatile solute is ?0.072°C. What would you expect the normal boiling point of this same solution to be? have no clue, thanks for the thelp Freezing point depression= N*kf*
1) For a freezing point depression experiment we had to create an ice bath and add salt to the ice. I have to answer the question, why was salting the ice necessary in order to complete the experiment and how does this apply to
The equation for lowering the freezing point of a solvent is given in your manual. Given that the freezing point for the pure solvent is 79.4 °C , the molality is 0.1 m , and the freezing point depression constant is 6.9 °C/m ,
Experiment questions: Experiment 1 results mass of beaker:116.944 g mass of dodecanoic:11.629 g hot water: 175 C warm water: 10 C Average Fressing pt of Dodecanoic = 42.7 Experiment 2 results: Unknown C solution mass: 2.091g Hot
a. Calculate the expected freezing-point depression of a 0.200 m KNO3 solution. b. Will the value you calculated match the actual freezing-point depression for this solution? Why or why not? a). -0.372 degrees Celsius b). Possibly
Calculate the freezing point of a solution containing 12.2grams of benzoic acid, C6H5CO2H, dissolved in 250 grams of nitrobenzene, C6H5NO2. The freezing point of nitrobenzene is 7.7 C, and its freezing point depression constant is