Quantitative Chemical Analysis
Quantitative Chemical Analysis
9th Edition
ISBN: 9781464135385
Author: Daniel C. Harris
Publisher: W. H. Freeman
Question
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Chapter 2, Problem 2.AE

(a)

Interpretation Introduction

Interpretation:

The true mass of water at 15°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,

m=m'(1-dadw)(1-dad)

Where,

m is mass of object in vacuum.

m' is mass of object in air (read on a balance).

da is the density of air (0.001 2 g/mL near I bar and 25°C) .

dw is the density of calibration weights (8.0 g/mL) .

d is the density of the object being weighed.

(a)

Expert Solution
Check Mark

Answer to Problem 2.AE

The true mass of water at 15°C is 5.4031g

Explanation of Solution

To calculate true mass of water at 15°C

Given,

Mass of water in air (balance reading) is 5.397 4 g

The density of air is 0.001 2 g/mL

Density of balance weight is 8.0 g/mL

We know,

Density of water at 15°C is 0.999 102 6 g/mL

True massofWater=(5.3974g)(1-0.0012g/mL8.0g/mL)(1-0.0012g/mL0.991026g/mL)=5.4031g

The given mass of water, density of water at 15°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 15°C is 5.4031g

Conclusion

The true mass of water at 15°C was calculated.

(b)

Interpretation Introduction

Interpretation:

The true mass of water at 25°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,

m=m'(1-dadw)(1-dad)

Where,

m is mass of object in vacuum.

m' is mass of object in air (read on a balance).

da is the density of air (0.001 2 g/mL near I bar and 25°C) .

dw is the density of calibration weights (8.0 g/mL) .

d is the density of the object being weighed.

(b)

Expert Solution
Check Mark

Answer to Problem 2.AE

The true mass of water at 25°C is 5.4031g

Explanation of Solution

To calculate true mass of water at 25°C

Given,

Mass of water in air (balance reading) is 5.397 4 g

The density of air is 0.001 2 g/mL

Density of balance weight is 8.0 g/mL

We know,

Density of water at 25°C is 0.997 047 9 g/mL

True massofWater=(5.3974g)(1-0.0012g/mL8.0g/mL)(1-0.0012g/mL0.997047 9 g/mL)=5.4031g

The given mass of water, density of water at 15°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 25°C is 5.4031g

Conclusion

The true mass of water at 25°C was calculated.

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