Answered: What is the freezing point of 2.47 m…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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What is the freezing point of 2.47 m solution of Napthalene in Benzene? Kf= 5.12° C/ms The freezing point og Benzene is 5.5°C

Expert Solution

Step 1

Solution is made up of two components which are solute and solvent. Solute is present in the minor amount and solvent is present in the major amount. Here solute is naphthalene and solvent is benzene.

Step 2

Write the expression of depression in the freezing point of a solution.

$∆ T = K_{f} m$

where m is the molality, K_f is the molal depression constant and $∆ T$ is the depression in freezing point.

The value of molal depression constant is 5.12^oC/m.

Molality of benzene is 2.47 m.

Substitute 5.12^oC/m for K_f and 2.47 m for m to calculate the depression in freezing point of solution.

$∆ T = K_{f} m$

So, the depression in freezing point of solution is $\begin{array}{rcl} 12 \end{array} \begin{array}{rcl} . \end{array} \begin{array}{rcl} 65 \end{array} \begin{array}{rcl} \end{array}$ ^oC.

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