   # If 52.5 g of LiF is dissolved in 306 g of water, what is the expected freezing point of the solution? (Assume the van’t Hoff factor, i , for LiF is 2.) ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 13, Problem 47PS
Textbook Problem
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## If 52.5 g of LiF is dissolved in 306 g of water, what is the expected freezing point of the solution? (Assume the van’t Hoff factor, i, for LiF is 2.)

Interpretation Introduction

Interpretation: The expected freezing point of LiFsolution has to be determined.

Concept introduction:

Colligative properties: Properties of solutions which having influence on the concentration of the solute in it. Colligative properties are,

• Decrease in the vapor pressure
• Increase in the boiling point
• Decline in the freezing point
• Osmotic pressure

Decline in the freezing point is huge when solute is an electrolyte than when solute is nonelectrolyte. Therefore, change in freezing point is calculated by using the equation,

ΔTfp=Kfpmsolutei

where,

Kfp is the molal freezing point depression constant.

i is van’t Hoff factor

van’t Hoff factor, i: it is the relation between change in in freezing point measured and change in in freezing point calculated. It indicates the total number of ions that are produced.

Molality (m): Molality is the number of moles of solute present in one kilogram of solvent.

Molality (m) =Numberofmolesofsolute1kgofsolvent

### Explanation of Solution

Given,

i=2

Mass of LiFis 52.5g

Mass of water is 306g=0.306kg

Molal freezing point depression constant of water is 1.860C/m

The number of moles of any substance can be determined using the equation

Numberofmole=GivenmassofthesubstanceMolarmass=52.5g25.94g/mol=2.02mol

The molality of LiFis,

Molality (m)=Numberofmolesofsolute1kgofsolvent=2

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