Pure benzophenone freezes at 48.1 degree C. A solution of 1.25 g of biphenyl (C_

Question

Pure benzophenone freezes at 48.1 degree C. A solution of 1.25 g of biphenyl (C_12H_10; molar mass = 154.2 g/mol) plus 25.38 g of benzophenone froze at 16.8 degree C. Calculate the K_f of benzophenone. A mixture of 1.03 g of unknown plus 23.94 g of benzophenone had the following cooling behavior: (a) Prepare a graph showing temperature on the ordinate and time on the abscissa. Label the point at which crystals first appeared, and identify the freezing temperature of the solution. (b) Using the data for benzophenone from Problem 6, calculate the molality of the solution, then the molar mass of the unknown.

Explanation / Answer

6)
delta Tf = 48.1 oC - 16.8 oC = 31.3 oC
number of mol of solute = mass/molar mass
= 1.25 / 154.2
= 8.11*10^-3 mol

mass of solvent = 25.38 g = 0.02538 Kg

molality, mb = number of mol / mass of solvent
= (8.11*10^-3)/(0.02538)
= 0.32 molal

use:
delta Tf = Kf*mb
31.3 = Kf*0.32
Kf = 98 oC/m

I am allowed to answer only 1 question at a time

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