Chapter 12, Problem 125APP

(a)

Interpretation Introduction

Interpretation: The final volume of the solution should be calculated.

Concept Introduction: mass/volume percent (m/V) concentration is defined as the mass of solute in grams for 100 mL of the solution. Here, mass of solute should be in grams and unit of volume of solution should be in millilitres.

Mathematically,

  $\begin{array}{l}\text{mass /volume percent}\left(m/V\right)=\frac{\text{mass of solute}\left(g\right)}{\text{volume of solute}\left(\text{mL}\right)+\text{volume of solvent}\left(\text{mL}\right)}\times 100\%\\ =\frac{\text{mass of solute}\left(g\right)}{\text{volume of solution}\left(\text{mL}\right)}\times 100\%\end{array}$

(b)

Interpretation Introduction

Interpretation: The final volume of the solution should be calculated.

Concept Introduction: Molarity is defined as the amount of solution present in 1 L of the solution. It is mathematically represented as follows:

  $M=\frac{n\left(\text{solute}\right)}{V\left(\text{solution,L}\right)}$

Here, n is number of moles and V is volume of solution in L.

The number of moles is related to mass and molar mass as follows:

  $n=\frac{m}{M}$

Here, m is mass and M is molar mass.

(c)

Interpretation Introduction

Interpretation: The final volume of the solution should be calculated.

Concept Introduction: Molarity is defined as the amount of solution present in 1 L of the solution. It is mathematically represented as follows:

  $M=\frac{n\left(\text{solute}\right)}{V\left(\text{solution,L}\right)}$

Here, n is number of moles and V is volume of solution in L.

The number of moles is related to mass and molar mass as follows:

  $n=\frac{m}{M}$

Here, m is mass and M is molar mass.