If an 0.860 m aqueous solution freezes at –2.10 °C, what is the van\'t Hoff factor, i, of the solute? Kf values can be found here.
depression in freezing ponit =kf*m*i
kf= Freezing point depression constant= 1.86 0degc/m
m= molality =0.86
2.1 = 1.86*0.86*i where i = Van't Hoff factor
i= 2.1/(1.86*0.86)=1.3. This can be approximated to 1
If an 0.860 m aqueous solution freezes at –2.10 °C, what is the van\'t Hoff factor,...
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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...
For an aqueous solution of HF, determine the van't Hoff factor assuming 0% ionization. For the same solution, determine the van't Hoff factor assuming 100% ionization. A solution is made by dissolving 0.0300 mol HF in 1.00 kg of water. The solution was found to freeze at -0.0644°C. Calculate the value of i and estimate the percent ionization of HF in this solution.
van't Hoff Factor
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