Chapter 4.5, Problem 2PE

Interpretation Introduction

Interpretation:

The volume of $0.315M$$NaOH$ solution that contains $6.22g$ has to be calculated.

Concept Introduction:

 The mass of the precipitate is depends on the reactant masses.

 Mole: The mole of the solute is calculated by multiplication of concentration of solution and volume of the solution and it is,

 $\text{Mole}\text{=}\text{Concentration}\text{(M)×}\text{volume}\text{(L)}$

 $\text{Molarity = }\frac{\text{No}\text{. of moles of solute}}{\text{Volume of solution in L}}$

 Mass:

 The mass of solute present in the solution is given by multiplication of mole and molar mass of the solute.

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