Chapter 12, Problem 12.63P

Interpretation Introduction

Interpretation:

Volume of $\text{0}{\text{.865 M CaCl}}_2$ required to precipitate $\text{35}\text{.0 mL}$ of $\text{0}{\text{.500 M AgNO}}_{\text{3}}$ and amount of $\text{AgCl}$ thus produced has to be calculated.

Concept Introduction:

The formula to evaluate volume from molarity is given as follows:

  $V=\frac{n}{M}$

Here,

$M$ denotes molarity.

$n$ denotes number of moles.

$V$ denotes volume.