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UnbelievableDreams
 one year ago
I need help with Chemistry
UnbelievableDreams
 one year ago
I need help with Chemistry

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UnbelievableDreams
 one year ago
Best ResponseYou've already chosen the best response.0What 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
 one year ago
Best ResponseYou've already chosen the best response.0I know that molality is 0.41 but how do I get freezing and boiling point?

taramgrant0543664
 one year ago
Best ResponseYou've already chosen the best response.0This 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
 one year ago
Best ResponseYou've already chosen the best response.0I used this equation but I kept getting wrong answer. Can you show mw how?

taramgrant0543664
 one year ago
Best ResponseYou've already chosen the best response.0What numbers are you using for your values?

UnbelievableDreams
 one year ago
Best ResponseYou've already chosen the best response.0Kb= 2.53 C/m Kr = 5.1 C/m

taramgrant0543664
 one year ago
Best ResponseYou've already chosen the best response.0Your first step is to convert your grams of naphthalene and benzene into moles

taramgrant0543664
 one year ago
Best ResponseYou've already chosen the best response.0So 2.20g *(1 mol/128.17g/mol)=? 42.2 * (1mol/78.11g/mol)=?

taramgrant0543664
 one year ago
Best ResponseYou've already chosen the best response.0Here 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
 one year ago
Best ResponseYou've already chosen the best response.0I am still confused.

JFraser
 one year ago
Best ResponseYou've already chosen the best response.0molality 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
 one year ago
Best ResponseYou've already chosen the best response.00.017 mol o naphthalene 0.0422 kg of benzene

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Good, 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
 one year ago
Best ResponseYou've already chosen the best response.0Can you show me how? I am still confused.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0decrease 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 C2.099 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
 one year ago
Best ResponseYou've already chosen the best response.0What 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how did you do? what is your pure solvent and what is the molality?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sorry I see the molality is correct 10.74 m

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what are your values of k for the water?

UnbelievableDreams
 one year ago
Best ResponseYou've already chosen the best response.0I 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now multiply the values of k times the molality

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Freezing point 0(1.86x10.74)= ??? Boiling point 100+ (0.512 x 10.74) = ?????

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the freezing point of the solution is below zero is a negative number. the boiling point is above 100 C
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