An aqueous salt solution is formed by adding 61.65 g Iron (III) nitrate (solute) to water (solvent). What mass (in g) of water is used if the freezing point of the solution is -11.8 oC.
Kf H2O = 1.86 oC/m
An aqueous salt solution is formed by adding 61.65 g Iron (III) nitrate (solute) to water...
What is the freezing point of a 1.327 m aqueous solution of aluminum nitrate, Al(NO3)3? The Kf of water is 1.86°C/m. ________°C
Freezing point depression can be used to experimentally determine the van't Hoff factor of a solute in solution. Given the data in the table, please answer the questions below and determine the "real" van't Hoff factor of the solute. Experimental Results Mass of solvent (water) Freezing point of water Freezing point depression constant (Kf) of water Mass of solution Freezing point of solution 8.515 g 0.00°C 1.86°C/m 9.3589 -5.45°C a. What mass of solute was used? b. What is the...
F. A solution was prepared by dissolving 0.800 g of sulfur S8, in 100.0 g of acetic acid, HC2H3O2. Calculate the freezing point and boiling point of the solution . For the solvent, Kb = 3.08 oC/m and Kf = 3.59 oC/m. G. Butylated hydroxytoluene (BHT) is used as an antioxidant in processed foods. (It prevents fats and oils from becoming rancid.) A solution of 2.500 g of BHT in 100.0 g of benzene had a freezing point of...
Determine the melting point of an aqueous solution containing 147 mg of saccharin (C7H5O3NS) added to 1.00 mL of water (density of water = 1.00 g/mL, Kf = 1.86°C/m). What is the freezing point of a 1.276 m aqueous solution of aluminum nitrate, Al(NO3)3? The Kfof water is 1.86°C/m.
A solution is made by dissolving 0.749 mol of nonelectrolyte solute in 861 g of benzene. Calculate the freezing point, Tf, and boiling point, Tb, of the solution. Constants may be found here. Solvent Formula Kf value* (°C/m) Normal freezing point (°C) Kb value (°C/m) Normal boiling point (°C) water H2O 1.86 0.00 0.512 100.00 benzene C6H6 5.12 5.49 2.53 80.1 cyclohexane C6H12 20.8 6.59 2.92 80.7 ethanol C2H6O 1.99 –117.3 1.22 78.4 carbon tetrachloride CCl4 29.8 –22.9 5.03 76.8...
A solution contains 0.180 g of an unknown nonelectrolyte solute in 50.0 g of water. The solution freezes at –0.040 0C (notice the negative sign). A) What is the molar mass of the unknown solute? B) What is the boiling temperature of the solution? Kf(water) = 1.86 K kg mol-1 ; Kb(water) = 0.51 K kg mol-1 . Assume that the solute only exists as a monomer in aqueous solution.
A student performing this experiment finds that 5.493 g of an unknown nonelectrolyte solute causes 42.588 g of water solvent to freeze at -1.56 oC. Knowing the freezing point of pure water is 0.00oC, calculate the molar mass of the unknown solute in g/mol. Kffor water is 1.86 oC/molal. What is mass of solvent in kilograms?
A solution is made by dissolving 0.585 mol of nonelectrolyte solute in 877 g of benzene. Calculate the freezing point, Tf, and boiling point, Tb, of the solution. Constants can be found in the table of colligative constants. Constants for freezing-point depression and boiling-point elevation calculations at 1 atm: Solvent Formula Kf value* (°C/m) Normal freezing point (°C) Kb value (°C/m) Normal boiling point (°C) water H2O 1.86 0.00 0.512 100.00 benzene C6H6 5.12 5.49 2.53 80.1 cyclohexane C6H12 20.8...
A solution was made by adding 800 g of ethanol, C2H5OH, to 8 x 103 g of water. How much would this lower the freezing point? Kf of H2O is 1.86 degrees Celsius/m The answer is supposed to be 4.1 degrees Celsius, but I got 3.98287 degrees Celsius. What did I do wrong?
A solution containing 1.00 g of an unknown non-electrolyte liquid and 9.00 g water has a freezing point of -3.33 oC. The Kf = 1.86 oC/m for water. Calculate the molar mass of the unknown liquid, in g/mol.