Chapter 9.6, Problem 9.17WE

Interpretation Introduction

Interpretation:

The molar mass of given monoprotic acid should be calculated.

Concept introduction:

Neutralization reaction:

• The reaction of acid and base to give salt is known as neutralization reaction.
• Equal volume and concentrations of acid react with equal volume of the base with same concentration.
• In end point the ionic concentration is zero.

Volumetric principle:

• The relationship between initial and final concentrations and volumes of solutions in titration process are given in the volumetric equation and it is,

$\begin{array}{l}{\text{M}}_{\text{c}}{\text{×mL}}_{\text{c}}\text{=}{\text{M}}_{\text{d}}{\text{×mL}}_{\text{d}}\\ {\text{M}}_{\text{c}}\text{=}\text{Initial}\text{concentration}\\ {\text{mL}}_{\text{c}}\text{=}\text{Initial}\text{volume}\\ {\text{M}}_{\text{d}}\text{=}\text{Final}\text{concentration}\\ {\text{mL}}_{\text{d}}\text{=}\text{Final}\text{volume}\end{array}$

Molarity:

The concentration of the solutions is given by the term of molarity and it is given by ratio between numbers of moles of solute present in litter of solution.

$\text{Molarity}\text{=}\frac{\text{No}\text{.mole}}{\text{volume}\text{(L)}}$

Mole:

The mole of compound is given by the ratio between taken mass of the compound and molar mass of the compound.

$\text{Mole}\text{=}\frac{\text{Mass}\text{of}\text{the}\text{compound}}{\text{Molar}\text{mass}\text{of}\text{the}\text{compound}}$

Molar mass:

The ration between taken mass of compound and number of moles in sample is to give a molar mass of the compound.

$Molarmass=\frac{Mass}{mole}$