Chapter 2, Problem 2.AE

(a)

Interpretation Introduction

Interpretation:

The true mass of water at $\text{15}°\text{C}$ should be calculated.

Concept introduction:

Buoyancy Correction:

The weight of the object in air is less then weight of the object in vacuum because of buoyancy, this weight variation is corrected using Buoyancy equation it is,

$\text{m}\text{=}\frac{\text{m'}\left(\text{1-}\frac{{\text{d}}_{\text{a}}}{{\text{d}}_{\text{w}}}\right)}{\left(\text{1-}\frac{{\text{d}}_{\text{a}}}{\text{d}}\right)}$

Where,

$\text{m}$ is mass of object in vacuum.

$\text{m'}$ is mass of object in air (read on a balance).

${\text{d}}_{\text{a}}$ is the density of air $\left(\text{0}\text{.001 2 g/mL near I bar and 25°C}\right)$.

${\text{d}}_{\text{w}}$ is the density of calibration weights $\text{(8}\text{.0 g/mL)}$.

$\text{d}$ is the density of the object being weighed.

Answer to Problem 2.AE

The true mass of water at $\text{15}°\text{C}$ is $\text{5}\text{.403}\text{1}\text{g}$

Explanation of Solution

To calculate true mass of water at $\text{15}°\text{C}$

Given,

Mass of water in air (balance reading) is $\text{5}\text{.397 4 g}$

The density of air is $\text{0}\text{.001 2 g/mL}$

Density of balance weight is $\text{8}\text{.0 g/mL}$

We know,

Density of water at $\text{15}°\text{C}$ is $\text{0}\text{.999 102 6 g/mL}$

$\begin{array}{l}\text{True mass}\text{of}\text{Water}\text{=}\frac{\text{(5}\text{.3974}\text{g}\text{)}\left(\text{1-}\frac{\text{0}\text{.0012}\text{g/}\text{mL}}{\text{8}\text{.0}\text{g/mL}}\right)}{\left(\text{1-}\frac{\text{0}\text{.0012}\text{g/mL}}{\text{0}\text{.991026}\text{g/mL}}\right)}\\ \text{=}\text{5}\text{.403}\text{1}\text{g}\end{array}$

The given mass of water, density of water at $\text{15}°\text{C}$, density of air, and density of balance weight are plugged in above equation to give true  mass of water.

The true mass of water at $\text{15}°\text{C}$ is $\text{5}\text{.403}\text{1}\text{g}$

Conclusion

The true mass of water at $\text{15}°\text{C}$ was calculated.

(b)

Interpretation Introduction

Interpretation:

The true mass of water at $\text{25}°\text{C}$ should be calculated.

Concept introduction:

Buoyancy Correction:

The weight of the object in air is less then weight of the object in vacuum because of buoyancy, this weight variation is corrected using Buoyancy equation it is,

$\text{m}\text{=}\frac{\text{m'}\left(\text{1-}\frac{{\text{d}}_{\text{a}}}{{\text{d}}_{\text{w}}}\right)}{\left(\text{1-}\frac{{\text{d}}_{\text{a}}}{\text{d}}\right)}$

Where,

$\text{m}$ is mass of object in vacuum.

$\text{m'}$ is mass of object in air (read on a balance).

${\text{d}}_{\text{a}}$ is the density of air $\left(\text{0}\text{.001 2 g/mL near I bar and 25°C}\right)$.

${\text{d}}_{\text{w}}$ is the density of calibration weights $\text{(8}\text{.0 g/mL)}$.

$\text{d}$ is the density of the object being weighed.

Answer to Problem 2.AE

The true mass of water at $\text{25}°\text{C}$ is $\text{5}\text{.403}\text{1}\text{g}$

Explanation of Solution

To calculate true mass of water at $\text{25}°\text{C}$

Given,

Mass of water in air (balance reading) is $\text{5}\text{.397 4 g}$

The density of air is $\text{0}\text{.001 2 g/mL}$

Density of balance weight is $\text{8}\text{.0 g/mL}$

We know,

Density of water at $\text{25}°\text{C}$ is $\text{0}\text{.997 047 9 g/mL}$

$\begin{array}{l}\text{True mass}\text{of}\text{Water}\text{=}\frac{\text{(5}\text{.3974}\text{g}\text{)}\left(\text{1-}\frac{\text{0}\text{.0012}\text{g/}\text{mL}}{\text{8}\text{.0}\text{g/mL}}\right)}{\left(\text{1-}\frac{\text{0}\text{.0012}\text{g/mL}}{\text{0}\text{.997047 9 g/mL}}\right)}\\ \text{=}\text{5}\text{.403}\text{1}\text{g}\end{array}$

The given mass of water, density of water at $\text{15}°\text{C}$, density of air, and density of balance weight are plugged in above equation to give true  mass of water.

The true mass of water at $\text{25}°\text{C}$ is $\text{5}\text{.403}\text{1}\text{g}$

Conclusion

The true mass of water at $\text{25}°\text{C}$ was calculated.