   Chapter 13, Problem 77GQ

Chapter
Section
Textbook Problem

A 2.0% (by mass) aqueous solution of novocainium chloride (C13H21ClN2O2) freezes at −0.237 °C. Calculate the van’t Hoff factor, i. How many moles of ions are in the solution per mole of compound?

Interpretation Introduction

Interpretation: van’t Hoff factor has to be calculated. The number of moles of ions produced in the solution per mole of compound 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

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

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

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

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

Numberofmole=GivenmassofthesubstanceMolarmass

Explanation

Given,

ΔTfp=0.2370CKfpofwater=1.860C/m

Mass of novocainium chloride is 2100×100g=2g

Molar mass of novocainium chloride is 272.77g/mol

Mass of water is 100g2g=98g

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

Numberofmole=GivenmassofthesubstanceMolarmass=2g272.77g/mol=7.33×103mol

Molality of the novocainium chloride solution is,

Molality (m) =Numberofmolesofsolute1kgofsolvent=7

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