A 0.194-g sample of a nonvolatile solid solute dissolves in 9.82 g of cyclohexan

Question

A 0.194-g sample of a nonvolatile solid solute dissolves in 9.82 g of cyclohexane. The change in the freezing point of the solution is 2.94 degrees Celsius.

From the talbe: Kf = 20

a. What is the molality of the solute in the solution?
I got .147 but I didn't think this seemed right.

b. Calculate the molar mass of the solute

c. The same mass of solute is dissolved in 9.82 g of t-butanol instead of cyclohexane. What is the fexpected freezing point change of the solution?

From the table: t-Butanol Kf=9.1, Freezing point = 25.5

Explanation / Answer

kf = 20.0 2.94 = 20.0 x m m = 0.147 = moles solute / 9.82 x 10^-3 Kg moles solute = 0.00144 molar mass = 0.194 g/ 0.00144 mol=134.4 g/mol delta T = m x kb = 0.147 x 2.80=0.412

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