UnbelievableDreams
  • UnbelievableDreams
I need help with Chemistry
Chemistry
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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UnbelievableDreams
  • UnbelievableDreams
What is the (a) molality, (b) freezing point, and (c) boiling point of a solution containing 2.20 g napthalene (C10H8) in 42.2 g of benzene (C6H6)?
UnbelievableDreams
  • UnbelievableDreams
I know that molality is 0.41 but how do I get freezing and boiling point?
marihelenh
  • marihelenh
@taramgrant0543664

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taramgrant0543664
  • taramgrant0543664
This is the equation for freezing point ΔT = i Kf m where T is the change in temp, i is the van't Hoff factor, Kf is the molal dreezing point depression constant and m is the molality of the solute
UnbelievableDreams
  • UnbelievableDreams
I used this equation but I kept getting wrong answer. Can you show mw how?
taramgrant0543664
  • taramgrant0543664
What numbers are you using for your values?
UnbelievableDreams
  • UnbelievableDreams
Kb= 2.53 C/m Kr = 5.1 C/m
taramgrant0543664
  • taramgrant0543664
Your first step is to convert your grams of naphthalene and benzene into moles
taramgrant0543664
  • taramgrant0543664
So 2.20g *(1 mol/128.17g/mol)=? 42.2 * (1mol/78.11g/mol)=?
taramgrant0543664
  • taramgrant0543664
Here I'm sorry I have to go but I found a link that goes through the steps for you Im sorry but I hope it's clear. http://chemed.chem.purdue.edu/genchem/probsolv/colligative/kf1.3.html
UnbelievableDreams
  • UnbelievableDreams
I am still confused.
UnbelievableDreams
  • UnbelievableDreams
@JFraser
JFraser
  • JFraser
molality is moles of solute per kilogram of solvent, so convert the 2.20g of naphthalene into moles, and the 42.2g of benzene into kilograms
UnbelievableDreams
  • UnbelievableDreams
0.017 mol o naphthalene 0.0422 kg of benzene
Cuanchi
  • Cuanchi
Good, you already calculate the molality and you said it is 0.41 m you also need the freezing and boiling temperature of the pure solvent (benzene) then with the formula ΔT = Kf m you calculate how much the temperature is going to decrease (k=5.12) (freezing point) or increase (k=2.53) (boiling point) and subtract or add that values from the values of freezing and boiling the pure solvent
UnbelievableDreams
  • UnbelievableDreams
Can you show me how? I am still confused.
Cuanchi
  • Cuanchi
decrease of freezing point ΔT = Kf m =5.12 C/m x 0.41 m = 2.099 C Freezing point of pure benzene = 5.5 C Freezing point of the solution 5.5 C-2.099 C = 3.452 C for the boiling point use the same equation ΔT = Kf m but use the other k=2.53 C/m Boiling point of pure benzene = 80.1C Boiling point of the solution 80.1C + (2.53 C/m x 0.41 m)= ???
UnbelievableDreams
  • UnbelievableDreams
What is the (a) molality, (b) freezing point, and (c) boiling point of a solution containing 100.0 g of ethylene glycol (C2H6O2) in 150.0 g of water? M= 10.74 How about this? I tried using this equation and it is wrong answer.
Cuanchi
  • Cuanchi
how did you do? what is your pure solvent and what is the molality?
Cuanchi
  • Cuanchi
sorry I see the molality is correct 10.74 m
Cuanchi
  • Cuanchi
what are your values of k for the water?
UnbelievableDreams
  • UnbelievableDreams
I understood the first question. I have another which I am confused. m= 10.74 Kb for water = 0.512 C/m Kf for water = 1.86 C/m
Cuanchi
  • Cuanchi
now multiply the values of k times the molality
Cuanchi
  • Cuanchi
Freezing point 0-(1.86x10.74)= ??? Boiling point 100+ (0.512 x 10.74) = ?????
Cuanchi
  • Cuanchi
the freezing point of the solution is below zero is a negative number. the boiling point is above 100 C

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