What is the freezing point of solution containing 6.78 grams Naphthalene (molar mass is 128.2 g/mol) dissolved in 16.0 grams paradichlorobenzene? the freezing point of pure paradichlorobenzene is 53.0 degrees C/ The freezing point depression constant is 7.10 degrees C/molal ( I think that the 7.10 would be Kf??? right)
Please explain how to figure this. ..?? Thanks!
2007-03-05 15:01:13 · 1 answers · asked by ~~Shelly~~ 2 in Science & Mathematics ➔ Chemistry
Yes, 7.10 is the Kf
If Tfo is the freezing point of pure solvent and Tf of the solution then the depression of the freezing point is
ΔTf = Tfo-Tf =Kf*m
m is the molality (moles of solute per kg of solvent)
but moles of solute =mass solute/MW (solute)
so if we express the mass of solvent also in grams we have
m= 1000*mass(solute) / (MW(solute) *mass(solvent) )
1000 comes from the fact that by expressing mass(solvent) in g we find mole solute/g solvent but we want mole solute/kg solvent
thus we need to muliply by 1000 to get molality
So our equation becomes
Tf = Tfo -Kf*1000*mass(solute) /(MW(solute)*mass(solvent) =
= 53.0 -7.1*1000*6.78/(128.2*16) = 29.5 deg C
2007-03-05 21:23:54 · answer #1 · answered by bellerophon 6 · 0⤊ 0⤋