Chapter U4, Problem 8RE

Interpretation Introduction

Interpretation: The steps that are followed to prepare $\text{1}\text{.0 M}$ silver nitrate needs to be described.

Concept Introduction: The molar concentration of a solute in a solution can be expressed in terms of molarity as moles of solute present in per liter of solution. Mathematically,

  $\text{Molarity = }\frac{\text{number of moles of solute}}{\text{Volume of solution (L)}}$ - (1)

Answer to Problem 8RE

The steps that are followed to prepare $\text{1}\text{.0 M}$ silver nitrate are:

• Weighing $\text{169}\text{.87 g}$ of silver nitrate.
• Adding the weighed amount of silver nitrate to 1 Liter volumetric flask.
• Filling the flask to the mark with distilled water.

Explanation of Solution

The solution to prepare is $\text{1}\text{.0 M}$ silver nitrate. The $\text{1}\text{.0 M}$ represents the concentration of silver nitrate in terms of molarity that means moles of silver nitrate in per liter of solution.

The moles of silver nitrate is calculated using formula:

  $\text{number of moles = }\frac{\text{mass of the substance}}{\text{Molar mass of the substance}}$

Molar mass of silver nitrate is $\text{169}\text{.87 g/mol}$ and for 1.0 M solution, the number of moles is 1. Substituting the values in order to calculate mass of silver nitrate is as:

  $\begin{array}{l}\text{1 mole = }\frac{\text{mass}}{\text{169}\text{.87 g/mol}}\\ \text{mass = 169}\text{.87 g/mol}\times \text{1 mole}\\ \text{mass = 169}\text{.87 g}\end{array}$

So, the required mass of silver nitrate is $\text{169}\text{.87 g}$ .

Hence, the steps that are followed to prepare $\text{1}\text{.0 M}$ silver nitrate are:

• Weighing $\text{169}\text{.87 g}$ of silver nitrate.
• Adding the weighed amount of silver nitrate to 1 Liter volumetric flask.
• Filling the flask to the mark with distilled water (1 L).