Chapter 8, Problem 8.80P

Interpretation Introduction

(a)

Interpretation:

The grams of solute in 250 mL of a 0.55 M ${\text{NaNO}}_{\text{3}}$ solution should be determined.

Concept Introduction:

Solution stoichiometry involves the calculation of the concentration of solutions in the given conditions of volumes, moles, etc. It is mainly based on the calculation of moles and volumes. These two values are used to calculate the molarity of a solution. The relation between moles, volume and molarity are as given below:

  $\text{Molarity = }\frac{\text{Moles of solute}}{\text{volume of solution in L}}$

  $\text{Moles = }\frac{\text{Mass of solute in g}}{\text{Molar mass of solute (g/mol)}}$

Interpretation Introduction

(b)

Interpretation:

The grams of solute in 145 mL of a 4.0 M ${\text{HNO}}_{\text{3}}$ solution should be determined.

Concept Introduction:

Solution stoichiometry involves the calculation of the concentration of solutions in the given conditions of volumes, moles, etc. It is mainly based on the calculation of moles and volumes. These two values are used to calculate the molarity of a solution. The relation between moles, volume and molarity are as given below:

  $\text{Molarity = }\frac{\text{Moles of solute}}{\text{volume of solution in L}}$

  $\text{Moles = }\frac{\text{Mass of solute in g}}{\text{Molar mass of solute (g/mol)}}$

Interpretation Introduction

(c)

Interpretation:

The grams of solute in 6.5 L of a 2.5 M $\text{HCl}$ solution should be determined.

Concept Introduction:

Solution stoichiometry involves the calculation of the concentration of solutions in the given conditions of volumes, moles, etc. It is mainly based on the calculation of moles and volumes. These two values are used to calculate the molarity of a solution. The relation between moles, volume and molarity are as given below:

  $\text{Molarity = }\frac{\text{Moles of solute}}{\text{volume of solution in L}}$

  $\text{Moles = }\frac{\text{Mass of solute in g}}{\text{Molar mass of solute (g/mol)}}$