Question & Answer: You dissolve 93.24 g of an unknown in 1000g of water and obtain deltaTf=2.34 degrees c…..

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 DeltaTf = 2.34oC

molal freezing point Kf=1.86 degrees c*kg/mol.

molality m= (given wt of solute/ gr mol wt)x (1000/Weight of solvent in gm)

DeltaTf = 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

Still stressed from student homework?
Get quality assistance from academic writers!