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You dissolve 93.24 g of an unknown in 1000g of water and obtain deltaTf=2.34 degrees c. What is the molecular weight of the unknown if Kf=1.86 degrees c*kg/mol?
Expert Answer
Answer
A. given depression in freezing point Tf = 2.34oC
molal freezing point Kf=1.86 degrees c*kg/mol.
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Question & Answer: You dissolve 93.24 g of an unknown in 1000g of water and obtain deltaTf=2.34 degrees c…..
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molality m= (given wt of solute/ gr mol wt)x (1000/Weight of solvent in gm)
Tf = Kf x molality
2.34 = 1.86 x (93.24/Mol wt) x (1000/1000) [given solvent weight = 1000g]
2.34 = 1.86 x (93.24/Mol wt)
Mol wt = 93.24×1.86 /2.34
Mol wt = 74.11gm
