Asked by Hannah
I need to calculate the molecular mass of ehtylene glycol based on the freezing point depression.
The teacher said that for glycol i=1 and to use the equation
m=delta T / Kf X i
Kf = -1.86C/m
so I did -2.5C / -1.86 X 1 = 1.344
I know that 1.344 can't be the weight for glycol but im not sure what to do now? Should i multiply by 1000?
The teacher said that for glycol i=1 and to use the equation
m=delta T / Kf X i
Kf = -1.86C/m
so I did -2.5C / -1.86 X 1 = 1.344
I know that 1.344 can't be the weight for glycol but im not sure what to do now? Should i multiply by 1000?
Answers
Answered by
DrBob222
1.344 is the molality.
Then molality = mols/kg solvent
Plug in kg solvent and m and solve for moles
Then mols = g/molar mass.
Rearrange to molar mass = g/mols. You know how much (grams) you used and you know mols, solve for molar mass.
Then molality = mols/kg solvent
Plug in kg solvent and m and solve for moles
Then mols = g/molar mass.
Rearrange to molar mass = g/mols. You know how much (grams) you used and you know mols, solve for molar mass.
Answered by
Hannah
the only info that I have is the mass of solute(g) and mass of solvent(kg) for glycol so im confused on what to use for mols?
Answered by
Hannah
Disregard question I understand now. I have to solve for mols.
Answered by
Hannah
so for molality I got 1.344 then I did
1.344 = mols/0.10kg = 0.1344 Im not sure if I did this step right
Then I did mm= g/mol
6.200 was the mass of the solute so
mm=6.200 / 0.1344 = 46.13.
I guess I made a mistake somewhere in my calculations because I thought that the mm of eythlene was around 62.
1.344 = mols/0.10kg = 0.1344 Im not sure if I did this step right
Then I did mm= g/mol
6.200 was the mass of the solute so
mm=6.200 / 0.1344 = 46.13.
I guess I made a mistake somewhere in my calculations because I thought that the mm of eythlene was around 62.
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