Chapter 2, Problem 2.12P

Interpretation Introduction

Interpretation:

Water was drained from a burette between the $\text{0}\text{.12 - 15}\text{.78 mL}$ marks.  The apparent volume was $\text{15}\text{.78-0}\text{.12 = 15}\text{.66mL}$, measured in air at ${\text{25 }}^{\text{o}}\text{C}$, the mass of water delivered was $\text{15}\text{.569 g}$.  The true volume has to be found.

Concept Introduction:

True mass:

The true mass of an object is a measure of its inherent inertia, and this is independent of buoyancy.  The true mass can be determined by weighing in a vacuum without buoyant forces.

The true mass can be determined by the formula:

    $\begin{array}{l}{\text{m = m}}^{\text{'}}\left(\text{1-}\frac{{\text{d}}_{\text{a}}}{{\text{d}}_{\text{w}}}\right)\text{/}\left(\text{1-}\frac{{\text{d}}_{\text{a}}}{\text{d}}\right)\\ \text{m = true mass}\\ {\text{m}}^{\text{'}}\text{= mass measured in air}\\ {\text{d}}_{\text{a}}\text{= density of air (0}{\text{.0012g/mL near 1bar and 25 }}^{\text{o}}\text{C)}\\ {\text{d}}_{\text{w}}\text{= density of balance weights (8}\text{.0g/mL)}\\ \text{d = density of object being weighed}\end{array}$