 # You dissolve 2.56 g of succinic acid, C 2 H 4 (CO 2 H) 2 , in 500. mL of water. Assuming that the density of water is 1.00 g/cm 3 , calculate the molality, mole fraction, and weight percent of acid in the solution. ### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781337399074 ### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781337399074

#### Solutions

Chapter
Section
Chapter 13, Problem 1PS
Textbook Problem

## You dissolve 2.56 g of succinic acid, C2H4(CO2H)2, in 500. mL of water. Assuming that the density of water is 1.00 g/cm3, calculate the molality, mole fraction, and weight percent of acid in the solution.

Expert Solution
Interpretation Introduction

Interpretation: The mole fraction, molality, and weight percent of succinic acid has to be identified.

Concept introduction:

Mole fraction: Amount of that component divided by the total amount of all of the components of the mixture ( nA +  nB +  n...)

Molality: The amount of solute (mol) per kilogram of solvent.

Concentration (C, mol/kg) =  molality of solute = Amount of solute (mol)Mass of solvent (Kg)

Weight percent: The mass of one component divided by the total mass of the mixture, multiplied by 100%

weight % A = Mass of A Mass of A + Mass of B + Mass of C + ...×100%

The calculated molality of succinic acid is 0.0434 m

The mole fraction of succinic acid is 0.000781

The weight percent of succinic acid is 0.509 %

### Explanation of Solution

• The Mole fraction:

Given:

Mass ofSuccinic acid = 2.56 gVolume of water = 500 ml

Massofwater=(500 ml)(1.00g/ml)=500g

Number of moles can be calculated by the equation given below:

No.ofmoles=massmolarmass

no.of moles (succinic acid) = 2.56 g 118.09 g/mol = 0.0217 molno.of moles (Water) = 500g 18 g/mol = 27.78 mol

Mole fraction can be determined using the general equation:

Mole fraction of A (χA)= nA nA +  nB +  n...

nsuccinicacid= 0.0217 molnWater= 27.78 molMole fraction of succinic acid: (χsuccinic acid) = nsuccinicacid nsuccinicacid +  nWater  = 0.0217 mol0.0217 mol + 27.78 mol = 0.000781

Convert the mass of succinic acid and water into unit of moles and substitute in the respective mole fraction equation as shown above. Hence, the mole fraction of succinic acid is 0.000781

• The molality:

Molality of solute = Amount of solute (mol)Mass of solvent (Kg)

nsuccinicacid = 0.0217 molAmount of solvent = 500 g

Molality is calculated as given below:

Amount of solute (mol)Mass of solvent (Kg)0.0217 mol0.500 kg = 0.0434 m

The molality of succinic acid is calculated as shown above. Hence, the calculated molality of succinic acid 0.0434 m

• The Weight percent:

Given:Mass ofSuccinic acid = 2.56 g

weight % A = Mass of A Mass of A + Mass of B + Mass of C + ...×100%

weight % succinic acid = Mass of succinic acid Mass of succinic acid + Mass of water ×100%2.56 g(2.56 g)+(500 g)×100%= = 0.509 %

The weight percent of the succinic acid is calculated as shown above. Hence, the weight percent obtained is 0.509 %

Conclusion

The mole fraction, molality, and weight percent of succinic acid were identified.

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