Boiling Point Elevation/Freezing Point Depression T = m i K Where: K b and K f d

Question

Boiling Point Elevation/Freezing Point Depression

T = m i K

Where:

Kb and Kf depend only on the SOLVENT. Below are some common values. Use these values for the calculations that follow.

Solvent

Formula

_Kb  (° C / m)

(°

*Please note that T as defined above will be a negative number. Therefore, Kf must also be given as a negative number.

You may also see T defined as the magnitude of the freezing point depression, a positive number. In this case, K f will also be a positive number. Regardless of the sign used for K f, remember that the freezing point of the solution is always depressed.

Problem:

The freezing point of benzene, C6H6, is 5.500 °C at 1 atmosphere. Kf(benzene) = -5.12 °C/m

In a laboratory experiment, students synthesized a new compound and found that when 13.13 grams of the compound were dissolved in 275.8 grams of benzene, the solution began to freeze at 4.427 °C. The compound was also found to be nonvolatile and a non-electrolyte.

What is the molecular weight they determined for this compound ?

____________g/mol

T = T(solution) - T(pure solvent) * m = (# moles solute / Kg solvent) i = (# moles of solute particles / mole of solute) Kb = boiling point elevation constant. Kf = freezing point depression constant.

Explanation / Answer

T = m i Kf

m = molality = moles of solute / mass of solvent = n / 0.2758

To - Tf = m x 5.12

5.5 - 4.427 = 5.12 m

m = 0.2096

molality = 0.2096 m

n / 0.2758 = 0.2096

n = 0.058

n = mass / molar mass

0.058 = 13.13 / molar mass

molar mass = 227.2 g/mol

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