   Chapter 13, Problem 65GQ

Chapter
Section
Textbook Problem

Instead of using NaCl to melt the ice on your sidewalk you decide to use CaCl2. If you add 35.0 g of CaCl2 to 150. g of water, what is the freezing point of the solution? (Assume i = 2.7 for CaCl2.)

Interpretation Introduction

Interpretation: The expected freezing point of CaCl2 is 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 ration between change in in freezing point measured and change in in freezing point calculated. It indicates the total number of ions that are produced.

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

Numberofmole=GivenmassofthesubstanceMolarmass

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

Molality (m) =Numberofmolesofsolute1kgofsolvent

Explanation

The expected freezing point of CaCl2 is calculated.

Given,

i=2.7

Given mass of CaCl2 is 35.0g

Molar mass of CaCl2 is 110.98g/mol

Mass of water is 150g=0.150kg

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=35.0g110.98g/mol=0.31537mol

The molality of CaCl2 is,

Molality (m)=Numberofmolesofsolute1kgofsolvent=0

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