   # The concentration of acetic acid, CH 3 CO 2 H, in a sample of vinegar is 5.0 weight percent. What is the concentration of vinegar expressed as mol/kg solvent (molality)? (a) 0.83 mol/kg (b) 0.88 mol/kg (c) 52 mol/kg (d) 0.083 mol/kg ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 13.1, Problem 2RC
Textbook Problem
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## The concentration of acetic acid, CH3CO2H, in a sample of vinegar is 5.0 weight percent. What is the concentration of vinegar expressed as mol/kg solvent (molality)? (a) 0.83 mol/kg (b) 0.88 mol/kg (c) 52 mol/kg (d) 0.083 mol/kg

Interpretation Introduction

Interpretation: The Molality and molality of vinegar is to be determined from the given option.

Concept introduction:

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

Numberofmole=GivenmassofthesubstanceMolarmass

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%

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

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

### Explanation of Solution

Reason for correct option

Given,

Molar mass of acetic acid is 60.05 g/mol

Weight percentage of vinegar is 5.0%

Weight percentage of vinegar indicates that 100g of vinegar contains 5g of acetic acid.

Mass of solvent is 100g5g=95g=0.095kg

The number of moles of acetic acid is using the equation

Numberofmole=5g60.05g/mol=0

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