Chapter 4, Problem 34PE

Interpretation Introduction

Interpretation:

The volume (in milliliters) of $\text{ 0}\text{.124 }M\text{ NaOH}$ that contains enough $\text{NaOH}$ to react with ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ in $\text{15}\text{.4 mL of 0}\text{.108 }M{\text{ H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is to be calculated.

Concept Introduction:

The number of moles present in one liter of solution is known as the molarity of the solution. It is calculated as follows:

$M=\frac{n}{V}$

Here, $M$ is the molarity, $V$ is the volume of solution in liters and $n$ is the number of moles of solute.

The relation between liter (L) and milliliter (mL) is given as: 1L = 1000 mL.