   # You make 20.0 g of a sucrose (C 12 H 22 O 11 ) and NaCl mixture and dissolve it in 1.00 kg water. The freezing point of this solution is found to be −0.426°C. Assuming ideal behavior, calculate the mass percent composition of the original mixture, and the mole fraction of sucrose in the original mixture. ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 10, Problem 121CP
Textbook Problem
143 views

## You make 20.0 g of a sucrose (C12H22O11) and NaCl mixture and dissolve it in 1.00 kg water. The freezing point of this solution is found to be −0.426°C. Assuming ideal behavior, calculate the mass percent composition of the original mixture, and the mole fraction of sucrose in the original mixture.

Interpretation Introduction

Interpretation:

The mass percentage of original mixture and the mole fraction of Sucrose have to be calculated.

Concept Introduction:

Mass percentage of a compound can be defined by as calculated mass of the compound to the total mass of the compound whole multiplied by 100 . The mass percentage of a compound can be given by the equation,

Masspercentage(%)=Calculatedmass(g)Totalmass(g)×100

Mole fraction of a compound can be defined as the number of moles of a substance to the total number of moles present in them. The mole fraction can be calculated by,

Molefractionofcompound=Numberofmoles(inmol)Totalnumberofmoles(inmol)

### Explanation of Solution

Record the given info

Mass of Sucrose and Sodium chloride = 20.0g

Mass of Water                                   = 1.00kg

Freezing point of the solution             = -0.426°C

To calculate the moles of solute

Molal depression freezing point of Water= 1.86°C/molal

From the change in freezing point depression in equation,

m =ΔTfKf=0.426°C1.86°C/molal=0.229molal

On assuming the solution density = 1.00gmL-1

Then, 0.229 moles of solute is present in 1.00L of the solution

NaClNa++Cl- i=2

Therefore,

2(molNaCl)+molC12H22O11=0.229mol

Moles of solute = 0.229mol

To calculate the mass of Sodium chloride and Sucrose

Molar mass of NaCl = 58.44 g

Molar mass of C12H22O11 = 342.3

Mass of NaCl + Mass of C12H22O11 = 20.0g

2nNaCl+nC12H22O11=0.22958.44(nNaCl)+342.3(nC12H22O11)=20.0Solvingtheaboveequation,nC12H22O11=0

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