Question
answer part 2 question b
show all work
Report 1. Graphs and calculations a. Plot temperature vers for each set of data. The temperature scale (the y axis us time be chosen so that it can be read to one decimal ing point of the solvent, the freezing point of On the graph, clearly label the freez solution and indicate the value for the freezing point depression b. c. Calculate the molality of the isopropyl alcohol-acetic acid solution d. From the known weights of solute and solvent, determine the GMW of isopropyl a (The density of glacial acetic acid is 1.05 g/ml) Report the accuracy of your result, expressed as % error. Answer (Show work for all numerical problems.) a. Predict which of the following would be the most effective anti-freeze, assuming 2. weights of each are used: methyl alcohol (CH3OH); sucrose (C12H2201i): ethyl (C2H5OH): ethylene glycol (C2H4(OH)2). Explain your choice. b. How would you prepare a 1.50 molal solution of sucrose in water? At what tem would the solution begin to boil? c. When 8.40 grams of a non-electrolyte was dissolved in 80.0 grams of water, the solution froze at -1.52 °C, Calculate the molecular weight of the solute. How many grams of ethylene glycol must you add to 10.00 liters of water in your car radiator to protect against freezing at-10.00 °C? To what Fahrenheit temperature does this d. correspond? solvent: no supercooling solvent: supercooling ip ime ime solution: no supercooling solution: supercooling fp time 20
Explanation / Answer
Part 2 question b
molality is the number of moles of solute per kg of the solvent.
molar mass of sucrose = 342.29 g/mol
We have to prepare 1.5 molal solutions,
For that 1.5 moles of sucrose is needed for 1 kg water.
Mass of sucrose required = number of moles x molar mass = 1.5 x 342.29 = 513.435 g i
Take 513.435 g and dissolve in 1000 g of water.
Ebullioscopic constant for water = 0.512 °C/molal
Elevation in boiling point = Ebullioscopic constant x molality = 0.512 x 1.5
Elevation in boiling poit = 0.768 °C
New boiling point = 100 + 0.768 = 100.768 °C.
