   Chapter 13, Problem 104SCQ

Chapter
Section
Textbook Problem

Sodium chloride (NaCl) is commonly used to melt ice on roads during the winter. Calcium chloride (CaCl2) is sometimes used for this purpose too. Let us compare the effectiveness of equal masses of these two compounds in lowering the freezing point of water, by calculating the freezing point depression of solutions containing 200. g of each salt in 1.00 kg of water. (An advantage of CaCl2 is that it acts more quickly because it is hygroscopic, that is. it absorbs moisture from the air to give a solution and begin the process. A disadvantage is that this compound is more costly.)

Interpretation Introduction

Interpretation:

The effectiveness of the given compounds NaClandCaCl2 in lowering the freezing point of water has to be determined by calculating the freezing point depressions of the solutions.

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

Freezing point depression: The freezing point of the solution varies with the solute concentration.

Freezing point depression = ΔTfp= Kfp. msolute,where,Kfp=molal freezing point depression constant,msolute= molality of solute.

Explanation

ΔTf=Kf×m×iwhereΔT=Changein freezingpoint Kfofwater=molal freezing point depression constant(-1.860c/m)m=Molality ofsolutioni=Van'tHofffactor

The freezing point depression of NaCl can be calculated.

Mass of NaCl is 200g

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

Numberofmole=GivenmassofthesubstanceMolarmass=200g58.4g/mol=3.425mol

Molality of the NaCl solution is,

Molality (m) =Numberofmolesofsolute1kgofsolvent=3.425mol1kg=3.425m

The freezing point depression of CaCl2 can be calculated.

Mass of CaCl2 is 200g

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

Numberofmole=GivenmassofthesubstanceMolarmass=200g110.98g/mol=1.8021mol

Molality of the CaCl2 solution is,

Molality (m) =Numberofmolesofsolute1kgofsolvent=1.8021mol1kg=1

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