Chapter 4, Problem 4.77QE

Interpretation Introduction

Interpretation:

The mass of Lead (II) sulfate precipitate has to be calculated, when mixing of $\text{20}\text{.0 mL}$ $\text{1}\text{.11}\text{M}$ solution of Lead (II) acetate with an excess of sodium sulfate solution.

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.

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