   Chapter 13, Problem 64GQ

Chapter
Section
Textbook Problem

The solubility of NaCl in water at 100 °C is 39.1 g/100. g of water Calculate the boiling point of this solution. (Assume i = 1.85 for NaCl.)

Interpretation Introduction

Interpretation: Boiling point of the solution 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

Boiling point elevation: The boiling point of the solution varies with the solute concentration.

Elevation in boiling point = ΔTbp= Kbp. msolute,where,Kbp=molal boiling point elevation constant,msolute= molality of solute,i=van't Hofffactor.

Increase in the boiling point is huge when solute is an electrolyte than when solute is nonelectrolyte.

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.

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

Molality (m) =Numberofmolesofsolute1kgofsolvent

Explanation

Boiling point of the solution is calculated.

Given,

Solubility of NaCl 39.1g in 100g of water

Molar mass of NaCl is 58.44g/mol

i=1.85

Molal boiling point elevation constant of water is 0.51210C/m

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

Numberofmole=GivenmassofthesubstanceMolarmass=39.1g54.44g/mol=0.669mol

The molality of the solute is,

Molality (m) =0.669mol0

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