Chapter 4, Problem 4.79QE

Interpretation Introduction

Interpretation:

The produced mass of solid precipitate has to be calculated, when on mixing $\text{10}\text{.0 mL of a 2}\text{.10 M}$ solution of Barium bromide 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.

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