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anonymous
 one year ago
solution is made by dissolving 25.5 grams of glucose (C6H12O6) in 398 grams of water. What is the freezingpoint depression of the solvent if the freezing point constant is 1.86 °C/m? Show all of the work needed to solve this problem.
anonymous
 one year ago
solution is made by dissolving 25.5 grams of glucose (C6H12O6) in 398 grams of water. What is the freezingpoint depression of the solvent if the freezing point constant is 1.86 °C/m? Show all of the work needed to solve this problem.

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taramgrant0543664
 one year ago
Best ResponseYou've already chosen the best response.1So for this one we have a couple extra steps so first we have to find the number of moles of glucose to do this: n=m/MM moles= mass / molar mass So mass is the 25.5g and the molar mass of glucose is 180.16 So 25.5/180.16=n n=?

taramgrant0543664
 one year ago
Best ResponseYou've already chosen the best response.1Great! So next step we need to find moles per kg of water but first we have to convert 398g into kg so you divide by 1000 as that is the conversion

taramgrant0543664
 one year ago
Best ResponseYou've already chosen the best response.1Yes so now we can figure out the moles per kg m=0.142/0.398 m=?

taramgrant0543664
 one year ago
Best ResponseYou've already chosen the best response.1Yes 0.357moles/kg, so the next step is the van't Hoff factor which again is represented by i, it is a dissociation factor, glucose doesn't dissociate in water so i=1 for this example

taramgrant0543664
 one year ago
Best ResponseYou've already chosen the best response.1Second last step now is to find our change in temperature: T=iKfm i is 1, Kf is given so 1.86 and m is the 0.357 we found previously

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0T=(1)(1.86)(0.357) = 0.664

taramgrant0543664
 one year ago
Best ResponseYou've already chosen the best response.1Yes so that is the change in temperature when the glucose is mixed in so to find the freezing point we add this to the freezing point of water So 0+(0.664)= 0.664 celcius is the freezing point of this reaction

taramgrant0543664
 one year ago
Best ResponseYou've already chosen the best response.1Because 0 is the freezing point of water

taramgrant0543664
 one year ago
Best ResponseYou've already chosen the best response.1When we did the boiling point we used 100 as that is the boiling point of water

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok so the answer is 0.664?

taramgrant0543664
 one year ago
Best ResponseYou've already chosen the best response.1Yes it would be

taramgrant0543664
 one year ago
Best ResponseYou've already chosen the best response.1No problem! Happy to help!!
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