Chapter 15, Problem 34PE

Interpretation Introduction

Interpretation:

Mass in grams of $\text{Al}{\left(\text{OH}\right)}_3$ required to neutralize with $275\text{ mL}$ of $\text{0}\text{.125 M HCl}$ has to be calculated.

Concept Introduction:

Molarity is quantitatively defined as moles of solute in one liter of solution. For example $\text{0}{\text{.070 M AlCl}}_3$ indicates that in $1\text{ L}$ the moles of ${\text{AlCl}}_3$ is $\text{0}\text{.070 mol}$. The formula to evaluate molarity is given as follows:

  $M=\frac{n}{V}$

Here,

$V$ represents volume.

$M$ represents molarity.

$n$ represents number of moles.

The formula to calculate mass from moles is given as follows:

  $\text{Mass}=\left(\text{Number of moles}\right)\left(\text{Molar mass}\right)$