Chapter 4, Problem 4.67QE

Interpretation Introduction

Interpretation:

The gram mass of $\text{AgCl}$ that is forms in the reaction of $\text{3}\text{.11 mL of 0}{\text{.11MAgNO}}_{\text{3}}$ with excess ${\text{CaCl}}_{\text{2}}$ has to be calculated.

Concept introduction:

Molarity: Dividing moles of solute by the volume of solution in Liter.

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

Mole: Dividing taken mass substance by molar mass of substance.

  $\text{Mole}\text{=}\frac{\text{Mass}}{\text{Molar}\text{mass}}$