Chapter 7, Problem 7.77AP

(a)

Interpretation Introduction

Interpretation:

The grams of acetic acid that is present in $\text{5}\text{.0\%}\left(\text{w/v}\right)$ of acetic acid in water has to be calculated.

Concept Introduction:

Weight/volume percent:  Weight/volume percent is defined as the concentration expressed in terms of grams of solute dissolved in volume of solution.  Weight/volume percent is given as,

  $\left(\text{w/v}\right)\text{\%=}\frac{\text{Mass}\text{of}\text{solute(in}\text{g)}}{\text{Volume}\text{of}\text{solution}\left(\text{in}\text{mL}\right)}\text{×100\%}$

(b)

Interpretation Introduction

Interpretation:

The moles of acetic acid that is present in $\text{5}\text{.0\%}\left(\text{w/v}\right)$ of acetic acid in water has to be calculated.

(c)

Interpretation Introduction

Interpretation:

The molarity has to be calculated from weight/volume percent of the solution.

Concept Introduction:

Molarity:  Molarity is defined as the mass of solute in one liter of solution.  Molarity is the preferred concentration unit for stoichiometry calculations.  The formula is,

  $\text{Molarity=}\frac{\text{Mass}\text{of}\text{solute(in}\text{moles)}}{\text{Volume}\text{of}\text{solution}\text{(in}\text{litres)}}$