Chapter 4, Problem 4.25P

(a)

Interpretation Introduction

Interpretation:

Molarity of the solution resulting from dissolving $46.0\text{ g}$ of silver nitrate in enough water to give a final volume of $335\text{ mL}$ is to be calculated.

Concept introduction:

Molarity $\left(M\right)$ is one of the concentration terms that determine the number of moles of solute present in per litre of solution. Unit of molarity is $\text{mol/L}$.

The expression to calculate the molarity of a solution when moles of solute and volume of solution are given is as follows:

$\text{Molarity of solution}\left(M\right)=\frac{\text{moles of solute}\left(\text{mol}\right)}{\text{volume of solution}\left(\text{L}\right)}$

The expression to calculate the mass of solute when moles and molecular mass of compound are given is as follows:

Molarity $\left(M\right)$ is one of the concentration terms that determine the number of moles of solute present in per litre of solution. Unit of molarity is $\text{mol}/\text{L}$.

The expression to calculate the molarity of a solution when moles of solute and volume of solution are given is as follows:

$\text{Molarity of solution}\left(M\right)=\frac{\text{moles of solute}\left(\text{mol}\right)}{\text{volume of solution}\left(\text{L}\right)}$

The expression to calculate the moles of solute when given mass and molecular mass of compound are given is as follows:

$\text{Moles of compound}\left(\text{mol}\right)=\left[\begin{array}{c}\text{given mass}\text{of compound}\left(\text{g}\right)\\ \left(\frac{\text{1mole}\text{of compound}\left(\text{mol}\right)}{\text{molecular mass of compound}\left(\text{g}\right)}\right)\end{array}\right]$

(b)

Interpretation Introduction

Interpretation:

The volume $\left(\text{L}\right)$ of $0.385M$$\text{manganese}\left(\text{II}\right)\text{sulfate}$ that contains $63.0\text{ g}$ of solute is to be calculated.

Concept introduction:

Molarity $\left(M\right)$ is one of the concentration terms that determine the number of moles of solute present in per litre of solution. Unit of molarity is $\text{mol}/\text{L}$.

The expression to calculate the volume of the solution when the amount of compound in moles and molarity of solution are given is as follows:

$\text{Volume of solution}\left(\text{L}\right)=\text{moles of solute}\left(\text{mol}\right)\left(\frac{\text{1}\text{L of solution}}{\text{molarity of solution}\left(\text{mol}\right)}\right)$

The expression to calculate the moles of solute when given mass and molecular mass of compound are given is as follows:

$\text{Moles of compound}\left(\text{mol}\right)=\left[\begin{array}{c}\text{given mass}\text{of compound}\left(\text{g}\right)\\ \left(\frac{\text{1mole}\text{of compound}\left(\text{mol}\right)}{\text{molecular mass of compound}\left(\text{g}\right)}\right)\end{array}\right]$

(c)

Interpretation Introduction

Interpretation:

The volume $\left(\text{mL}\right)$ of $6.44\times {10}^{-2}M$ adenosine triphosphate (ATP) that contains $1.68\text{ mmol}$ of ATP is to be calculated.

Concept introduction:

Molarity $\left(M\right)$ is one of the concentration terms that determine the number of moles of solute present in per litre of solution. Unit of molarity is $\text{mol}/\text{L}$.

The expression to calculate the volume of a solution when moles of solute and molarity of solution are given is as follows:

$\text{Volume of solution}\left(\text{L}\right)=\frac{\text{moles of solute}\left(\text{mol}\right)}{\text{Molarity of solution}\left(M\right)}$