Chapter 10, Problem 10.26P

Interpretation Introduction

Interpretation:

Molarity of ${\text{HNO}}_3$ when $\text{0}\text{.2619 g}$ of ${\text{Na}}_{\text{2}}{\text{B}}_{\text{4}}{\text{O}}_{\text{7}}$ requires to  $\text{21}{\text{.61 mL HNO}}_3$ for titration 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 $\text{0}\text{.070 mol}$.

Formula to calculate molar mass from number of moles is given as follows:

  $\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Formula to calculate molarity is as follows:

  $M=\frac{n}{V}$

Here,

$V$ denotes volume.

$M$ denotes molarity.

$n$ denotes moles.