f 78.1 g of a sodium tetrafluoroborate is dissolved in 465 g of water solvent, c
ID: 824220 • Letter: F
Question
f 78.1 g of a sodium tetrafluoroborate is dissolved in 465 g of water solvent, calculate the freezing temperature (if below 0 oC, include the sign) of the solution. Consider if the solute is an electrolyte.
If 1010 g of a calcium iodide is dissolved in 1230 g of water solvent, calculate the magnitude (not the sign) of the freezing point depression of the solution. Consider if the solute is an electrolyte.
If 37.7 g of a ethylene glycol is dissolved in 233 g of cyclohexane solvent, calculate the freezing temperature (if below 0 oC, include the sign) of the solution. Consider if the solute is an electrolyte.
If 408 g of a ethylene glycol is dissolved in 2650 g of benzene solvent, calculate the magnitude (not the sign) of the boiling point elevation of the solution. Consider if the solute is an electrolyte.
If 1130 g of a potassium citrate is dissolved in 2430 g of water solvent, calculate the freezing temperature (if below 0 oC, include the sign) of the solution. Consider if the solute is an electrolyte.
If 1160 g of a aluminum sulfate is dissolved in 2640 g of water solvent, calculate the magnitude (not the sign) of the boiling point elevation of the solution. Consider if the solute is an electrolyte.
Explanation / Answer
1)
DTf = i* Kf *molality
(0-Tf ) = 2*1.858*(78.1/109.79)(1000/465)
Tf = -5.684 0C
2)
DTf = i*Kf *molality
DTf = 3*1.858*(1010/293.89)1000/1230
DTf = 23.355
3)
DTf = i*Kf *molality
(0-Tf) = 3* 20*(37.7/62.068)(1000/233)
Tf = -106.39 0C
4)
DTb = i*Kb*molality
DTb = 3* 2.61(408/62.08)(1000/2650)
DTb = 12.946
5)
DTf = i*Kf *molality
(0-Tf) = 4*1.858*(1130/306.41)(1000/2430)
Tf = - 11.28 0C
6)
DTb = i*Kb*molality
(Tb-100) = 5*0.512*(1160/342.15)(1000/2640)
Tb = 103.285 0C
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